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CBSE Questions for Class 11 Medical Chemistry Hydrocarbons Quiz 13 - MCQExams.com
CBSE
Class 11 Medical Chemistry
Hydrocarbons
Quiz 13
$$[P]\overset{Br_2}{\longrightarrow}C_2H_4Br_2\overset{NaNH_2}{\underset{NH_3}{\longrightarrow}}Q$$
$$[Q] \overset{20\%H_2SO_4}{\underset{Hg^{2+}, \Delta}{\longrightarrow}}[R]\overset{Zn-Hg/HCl}{\longrightarrow}[S]$$.
The species P, Q, R and S respectively are ___________.
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Ethene, ethyne, ethanal, ethane
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Ethane, ethyne, ethanal, ethene
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Ethene, ethyne, ethanal, ethanol
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Ethyne, ethane, ethane, ethanol
Explanation
$$\underset{[P]}{CH_2=CH_2} \overset{Br_2}{\longrightarrow}Br-CH_2-CH_2-Br\overset{NaNH_2}{\underset{NH_3}{\longrightarrow}}\underset{[Q]}{HC\equiv CH}\overset{HgSO_4}{\underset{Hg^{+2}}{\longrightarrow}}\underset{[R]}{CH_3-CHO}\overset{zn-Hg}{\underset{HCl}{\longrightarrow}}\underset{[S]}{CH_3-CH_3}$$.
Product (A) is.
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In the given reactions:
(i) $$CH_3Br \xrightarrow [ ]{ Na,ether } X \xrightarrow [ hv ]{ Br_2 } Y \xrightarrow [ ]{ Na,ether } Z$$
(ii) $$CH_3COOH \xrightarrow [ ]{ NaOH } X \xrightarrow [ CaO ]{ NaOH } Y \xrightarrow [ hv ]{ Br_2 } Z$$
Identify $$X,\, Y$$ and $$Z$$.
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$$ (i) X= CH_4;\,\, Y= CH_3Br;\,\, Z= CH_3CH_3\,\,\,\,\,\,(ii) X = CH_3COONa; \,\, Y= CH_3CH_3;\,\, Z = CH_3CH_2Br$$
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$$ (i) X= CH_3CH_3;\,\, Y= CH_4;\,\, Z= CH_3Br\,\,\,\,\,\,\,(ii) X = CH_3COONa; \,\, Y= CH_3CH_3;\,\, Z = CH_3CH_2CH_3$$
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$$ (i) X= CH_3CH_2Br;\,\, Y= CH_3CH_3;\,\, Z= CH_3CH_2CH_3\,\,\,\,\,\,\,(ii) X = CH_3COONa; \,\, Y= CH_4;\,\, Z = CH_3Br$$
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$$ (i) X= CH_3CH_3;\,\, Y= CH_3CH_2Br;\,\, Z= CH_3CH_2CH_2CH_3\,\,\,\,\,\,\,(ii) X = CH_3COONa; \,\, Y= CH_4;\,\, Z = CH_3Br$$
Explanation
(i)
(a) When methyl bromide is treated with sodium metal in presence of dry ether, coupling takes place to give ethane. Reaction is known as Wurtz's reaction and also called as coupling reaction of alkyl halide.
$$C{H}_{3}Br \xrightarrow{Na, ether} C{H}_{3}-C{H}_{3}$$
Thus, $$X$$ is $$C{H}_{3}C{H}_{3}$$.
(b)The reaction of ethane with bromine called a substitution reaction.
$$C{H}_{3}-C{H}_{3} \xrightarrow[h \nu]{{Br}_{2}} C{H}_{3}C{H}_{2}Br$$
Thus, $$Y$$ is $$C{H}_{3}C{H}_{2}Br$$.
(c) Next reaction is also Wurtz's reaction and will form butane.
$$C{H}_{3}C{H}_{2}Br \xrightarrow{Na, ether} C{H}_{3}-C{H}_{2}-C{H}_{2}-C{H}_{3}$$
Thus, $$Z$$ is $$C{H}_{3}C{H}_{2}C{H}_{2}C{H}_{3}$$.
(ii)
(a)
When $$NaOH$$ added to acetic acid, t
hey react to form sodium acetate and water. This is a typical acid + base neutralization reaction.
$$C{H}_{3}COOH \xrightarrow{NaOH} C{H}_{3}COONa + {H}_{2}O$$
Thus, $$X$$ is $$C{H}_{3}COONa$$.
(b) When sodium acetate is treated with soda lime, decarboxylation takes place and $$CO_2$$ is eliminated from the sodium acetate.
$$C{H}_{3}COONa \xrightarrow[CaO]{NaOH} C{H}_{4} + {Na}_{2}C{O}_{3}$$
Thus, $$Y$$ is $$C{H}_{4}$$.
(c) This reaction between methane and bromine happens in the presence of ultraviolet light - typically sunlight. This is a good example of a photochemical reaction.
$$C{H}_{4} \xrightarrow[h \nu]{{Br}_{2}} C{H}_{3}Br + HBr$$
Thus, $$Z$$ is $$C{H}_{3}Br$$
In the above reaction rate is fastest, when (X) is:
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-OH
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$$NH_2$$
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$$- \overset{O}{\overset{||}{\underset{O}{\underset{||}{S}}}} - OCH_3$$
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$$- O - \overset{O}{\overset{||}{\underset{O}{\underset{||}{S}}}} - CH_3$$
Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make $$1^{\circ} ,2^{\circ} ,3^{\circ}$$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So ,bromine is less reactive, and more selective whereas chlorine is less selective
and more effective.
The relative rate of abstraction of hydrogen by Br is
$$\underset{(1600)}{3^\circ} \, > \, \underset{(82)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
The relative rate of abstraction of hydrogen by Cl is
$$\underset{(5)}{3^\circ} \, > \, \underset{(3.8)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
D.Which of the following will give five monochloro products, when allowed to react with $$Cl_2$$ in presence of sun light (excluding stereoisomers) ?
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n - pentane
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Iso - pentane
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2 - methyl - pentane
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3 - methyl - pentane
Explanation
To form five
monochloro
products refer the above image-
n-pentane has only three as refer to IMAGE $$01$$
Iso-pentane has only four as refer to IMAGE $$02$$ and
$$2-$$Methyl-pentane has five products as refer to the IMAGE $$03$$
$$CH_4+Cl_2\xrightarrow{hv} CH_3Cl+HCl$$
To obtain high yields of $$CH_3Cl,$$ the ratio of $$CH_4$$ to $$Cl_2$$ must be:
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high
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low
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equal
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can't be predicted
Explanation
$$CH_4+Cl_2 \xrightarrow [] {hv} CH_3Cl+HCl$$
To obtain high yields of $$CH_3Cl$$, the ratio of $$CH_4$$ to $$Cl_2$$ must be high, otherwise if we take low ratio of $$CH_4$$ to $$Cl_2$$ i.e. excess $$Cl_2$$ has been used then there will be formation of more than mono substituted products i.e. $$CH_2Cl_2, CHCl_3, CCl_4$$
An alkane $$C_{7}H_{16}$$ is produced by the reaction of lithium di(3-pentyl)cuprate with ethyl bromide. The name of the product is:
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3-methylhexane
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2-ethylpentane
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3-ethylpentane
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n-heptane.
Explanation
Lithium di(3-pentyl)cuprate is a Gilman reagent. It couples with alkyl halides to give alkanes.
$$\underset{\text{Lithium di(3-pentyl)cuprate}}{[{(C{H}_{3}C{H}_{2})}_{2}CH]_{2}CuLi} + \underset{\text{ethyl bromide}}{BrC{H}_{2}C{H}_{3}} \longrightarrow \underset{3-ethylpentane}{{C}_{2}{H}_{5}-\overset{\underset{|}{{C}_{2}{H}_{5}}}{CH}-{C}_{2}{H}_{5}} + {(C{H}_{3}C{H}_{2})}_{2}CHCu + LiBr$$
Hence the product is 3-ethylpentane.
Correct statement about Kolbe's electrolytic synthesis is:
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pH of reaction mixture increases as reaction goes on
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pH of reaction mixture decreases as reaction goes on
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pH of reaction mixture remain same as reaction goes on
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pH of reaction mixture first increases and that decreases.
Explanation
In Kolbe's electrolytic synthesis, an aqueous solution of sodium or potassium salt of a carboxylic acid on electrolysis give alkane.
The following reaction takes place:
$$2RCOO^-Na^++2H_2O\xrightarrow []{Electrolysis}R-R+2CO_2+H_2+2NaOH$$
As production of alkali i.e. $$NaOH$$ takes place that is why $$pH$$ of reaction mixture increases as reaction goes on.
Which of the following reactions will not give propane?
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$$CH_3CH_2CH_2Cl\overset{Mg/ether}{\underset{H_2O}{\rightarrow}}$$
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$$CH_3COCl\overset{CH_3MgX}{\underset{H_2O}{\rightarrow}}$$
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$$CH_3CH=CH_2\overset{B_2H_6}{\underset{CH_3COOH}{\rightarrow}}$$
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$$CH_3\underset{OH}{\underset{|}{C}}H-CH_3\overset{P/HI}{\rightarrow}$$
Explanation
$$(B)$$ will give the above image. So propane cannot be formed.
Sulphonation is most favourable at which carbon number?
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$$1$$
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$$2$$
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$$3$$
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$$4$$
Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make $$1^{\circ} ,2^{\circ} ,3^{\circ}$$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is
$$\underset{(1600)}{3^\circ} \, > \, \underset{(82)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
The relative rate of abstraction of hydrogen by Cl is
$$\underset{(5)}{3^\circ} \, > \, \underset{(3.8)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
F. What would be the product ratio x/y in the chlorination of propane if all the hydrogen were abstracted at equal rate ?
$$CH_3 - CH_2 - CH_3 \xrightarrow[hv]{Cl_2} CH_3 - CH_2 - \underset{(x)} CH_2 - Cl + CH_3 -\underset {(y)}{\underset{Cl} {\underset{|}{C}}}H - CH_3$$
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$$\dfrac{1}{3}$$
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$$\dfrac{3}{1}$$
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$$\dfrac{9}{1}$$
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$$\dfrac{1}{9}$$
Explanation
$$CH_3 - CH_2 - CH_3 \xrightarrow[hv]{Cl_2} CH_3 - CH_2 - \underset{(x)} CH_2 - Cl + CH_3 -\underset {(y)}{\overset{Cl} {\overset{|}{C}}}H - CH_3$$
We know, rate of reactivity is-
$$3^o>2^o>1^o$$
$$(5)$$ $$(3.8)$$ $$(1)$$
$$x\rightarrow 1^o\Rightarrow 6\times 1=6$$
$$y\rightarrow 2^o\Rightarrow 2\times 3.8=6.4$$
So, $$\cfrac {x}{y}=\cfrac {6.4}{6}=\cfrac {3.2}{3.0}$$
But, given rate is same.
So, $$x\rightarrow 1^o\Rightarrow 6\times 1=6$$
$$y\rightarrow 2^o\Rightarrow 2\times 1=2$$
$$\cfrac {x}{y}=\cfrac {6}{2}=\cfrac {3}{1}$$
Products obtained in above Wurtz reaction is:
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both $$(A)$$ and $$(B)$$
Explanation
Following dichlorocompound undergoes formation of diradicals:
(Refer to Image)
$$\therefore$$ Products will be both option A and B. So, Option D is the answer.
Arrange the compounds I, II and III in decreasing order of their heats of combustion:
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II > I > III
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I > II > III
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III > II > I
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III > I > II
Explanation
High ring strain will have less heat of combustion.
Order of heat of combustion: $$III > II > I$$, because ring strain is in the order $$I > II > III$$ i.e. $$5-$$membered ring strain is highest and as result it will have the lowest heat of combustion.
Arrange the following alkanes in decreasing order of their heats of combustion.
(i) $$\underset{(Neo-pentane) \, }{CH_3-\overset{CH_3}{\overset{|}{\underset{CH_3}{\underset{|}{C}}}}-CH_3}$$
(ii) $$\underset{(Iso-pentane)}{CH_3-\underset{CH_3}{\underset{|}{CH}}-CH_2-CH_3}$$
(iii)$$\underset{(n-pentane)}{CH_3-CH_2-CH_2-CH_2-CH_3}$$
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(i) > (ii) > (iii)
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(iii) > (I) > (ii)
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(iii) > (ii) > (i)
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(i) > (iii) > (ii)
Explanation
$$CH_3$$
$$H_3C-\underset {|}{\overset {|}{C}}-CH_3$$ $$CH_3-\underset {|}{CH}-CH_2-CH_3$$ $$CH_3CH_2CH_2CH_2-CH_3$$
$$CH_3$$ $$CH_3$$
$$Neopentane (i)$$ $$Iso-pentane (ii)$$ $$n-pentane (iii)$$
We know that alkanes that are more branched, have less heat of combustion than a less branch one. Thus here heat of combustion order will be:
$$n-pentane(iii) > isopentane (ii) > Neopentane (i)$$.
Because here $$Neopentane$$ is more branched than $$isopentane$$ which is again more branched than $$n-pentane$$.
Major product $$(A)$$ is:
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$$Ph - C \equiv C^{14} - Ph$$
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$$Ph - C \equiv C - Ph$$
Halogenation is a substitution reaction, where halogen replaces one or more hydrogens of hydrocarbon.
Chlorine free radical make $$1^{\circ} ,2^{\circ} ,3^{\circ}$$ radicals with almost equal ease, whereas bromine free radicals have a clear preference for the formation of tertiary free radicals. So , bromine is less reactive, and more selective whereas chlorine is less selective and more effective.
The relative rate of abstraction of hydrogen by Br is
$$\underset{(1600)}{3^\circ} \, > \, \underset{(82)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
The relative rate of abstraction of hydrogen by Cl is
$$\underset{(5)}{3^\circ} \, > \, \underset{(3.8)}{2^\circ} \, > \, \underset{(1)}{1^\circ}$$
B.Above product will obtained in better yield if X is
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$$Cl_2$$
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$$I_2$$
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$$Br_2$$
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Can't be predicted
Explanation
For the reaction $$R-H+X_2\overset {h\nu}{\longrightarrow} R-X+HX$$ to be obtained in better yeild, $$X$$ must be Bromine, This is because $$R$$ is a $$3^o$$ radical, hence we know relative rate of Bromination is very high.
On halogenation, an alkane gives only one monohalogenated product. The alkane may be:
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2-methyl butane
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2, 2-dimethyl propane
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cyclopentane
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both (b) and (c)
Explanation
A contains $$4$$ different $$H$$ atoms that's why A gives $$4$$ different mono halogenated product. But B and C have all equivalent $$H-$$atoms giving only one chlorinated (monosubstituted) product.
In the following reaction sequence the product B is:
$$Ph-C\equiv CH\xrightarrow [ aq.{ H }_{ 2 }{ SO }_{ 4 } ]{ { Hg }^{ 2+ } } A\xrightarrow [ 2.{ H }_{ 2 }O ]{ 1.{ CH }_{ 3}MgI } B$$
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$$Ph-CH_3-CH_2-OH$$
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$$Ph-(CH_3)_2C(OH)$$
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$$Ph-CH_3-COOH$$
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none of these
In the preparation of alkanes, a concentrated aqueous solution of sodium or potassium salts of saturated carboxylic acid are subjected to:
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hydrolysis
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oxidation
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hydrogenation
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electrolysis
Explanation
Carboxylic acids are electrolysed to form alkyl radical which combines to form alkane.
$$R-COO^- \rightarrow R-\overset{\overset{O}{||}}{C}-\dot{O}\rightarrow \dot{R}+CO_2$$
$$\dot{R}+\dot{R} \rightarrow R_2$$
Which of the following method can be used for the preparation of methane?
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Wurtz reaction
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Kolbe's reaction
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Reduction of alkyl halide
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Hydrogenation of alkene
Explanation
Among the given reaction, methane can only be prepared by reduction of alkyl halides.
All other methods prepare ethane as the smallest hydrocarbon.
The preparation of methane by reduction of alkyl halide can be given as follow:
$$CH_3-Cl \xrightarrow[H^+]{Zn}\underset{methane}{CH_4}$$
Hence, the correct option is $$(C)$$.
Which of the following systems is the correct proposition given by kekule regarding the stricture of benzene ?
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Explanation
The 2 kekule structures are drawn double headed resonance are between them as shown below.
This is meant to indicate these 2 structures are resonance structures of benzene and that taken together,they represent the true and physical structure of benzene.
Mesitylene is prepared from:
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$$C{H}_{3}CHO$$ and conc. $$HN{O}_{3}$$
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$$C{H}_{3}COC{H}_{3}$$ and conc. $${H}_{2}S{O}_{4}$$
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$$C{H}_{3}COC{H}_{3}$$ and conc. $$HCl$$
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$$C{H}_{3}CHO$$ and conc. $${H}_{2}S{O}_{4}$$
Explanation
Solution:- (B) $$C{H}_{3}COC{H}_{3}$$ and conc. $${H}_{2}S{O}_{4}; \quad \& \quad$$ (C) $$C{H}_{3}COC{H}_{3}$$ and conc. $$HCl$$
Mesitylene can be prepared by the action of sulfuric acid $$\left( {H}_{2}S{O}_{4} \right)$$ on acetone; by the action of hydrochloric acid on acetone, under pressure and at temperatures varying from $$100 ℃$$ to $$200 ℃$$.
$$C{H}_{3}-CO-C{H}_{3} \xrightarrow[-3 {H}_{2}O]{{H}_{2}S{O}_{4}/HCl} \text{Mesitylene}$$
How many number of $$1^{ \circ },\quad { 2 }^{ \circ }\quad and\quad { 3 }^{ \circ }$$ carbon atoms in the following compounds ?
$$CH_3-\underset{OH}{\underset{|\,\,\,\,}{CH}}-\underset{CH_3}{\underset{|\,\,\,\,}{CH}}-CH_2-CH_3$$
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3,2,2
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3,2,1
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3,1,2
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2,2,2
Explanation
No. of $$1^o$$ carbon$$-3$$
No. of $$2^o$$ carbon $$-1$$
No. of $$3^o$$carbon $$-2$$
$$-CH_3$$ is a$$1^o$$ carbon $$-CH_2-$$ as a type of $$2^o$$ carbon and $$-CH-$$ is a type $$-3^o-$$ carbon.
The barrier for rotation about the indicated bond is only $$14\ kcal/mol$$. This barrier is much lower than the barrier for rotation about $$C=C$$ bond in ethylene. The smaller barrier in this compound is due to:
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dipolar structure
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lack of resonance
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enolization
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combination of $$a$$ and $$b$$
How many ionizable hydrogen's are there in the following compound?
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2
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4
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5
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7
Explanation
Ionizable hyd are those where carbocation is stable:
Which of the following give alkene only by heating?
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What is number of electron delocalizing in benzene molecule?
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3
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6
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Zero
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12
Here $$X$$ is
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0%
0%
0%
Explanation
For the halogens, the electronegativity & electrophilicity decrease from $$F$$ to $$I$$ in the periodic table.
Flourine is most electrophilic & Iodine is least.
$$\therefore$$ Iodination is highly unreactive.
But when benzene is treated with iodine monochloride $$ICl$$ an interhalogen compound, aryl iodide is obtained. This is because, Chlorine is more electronegative & pulls the electron cloud giving iodine a slight positive charge.
The number of possible structural isomers for $$C_4H_8$$ are:
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$$4$$
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$$2$$
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$$6$$
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$$5$$
The number of different substitution products possible when ethane is allowed to react with bromine in sunlight are:
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$$7$$
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$$8$$
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$$9$$
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$$10$$
Which one of the following alksne is NOT formed in Wurtz reaction?
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Methane
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ethane
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propane
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butane
Explanation
Methane
In wurtz reaction a solution of alkyl halide in either on heating with sodium gives alkane
$$R-X+2Na+X-R\xrightarrow [ either ]{ Dry } R-R+2Na-X$$
An alkyl halide on wartz reaction leads to the formation of symmetrical alkane having an even of carbon atoms. Two different alkyl halides, on wartz reaction give all possible alkanes Hence, methane can't be prepared because it contains single c-atom.
The compound which contains all the four $${ 1 }^{ 0 },{ 2 }^{ 0 },{ 3 }^{ 0 }$$ and $$ { 4 }^{ 0 }$$ carbon atoms is
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2,3-dimethylpentane
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3-chloro-2,3-dimethylpentane
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2,3,4-trimethylpentane
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3,3-dimethylpentane
$$HC \equiv CH \xrightarrow{Excess \, Na} A \xrightarrow{Excess \, CH_3Cl} B$$.
Product $$B$$ is :
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2-butyne
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1-butyne
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propyne
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monosodium acetylide
Explanation
$$ HC \equiv CH \overset{Excess\, Na}{\rightarrow} Na -C\equiv C - Na$$
$$ Na -C \equiv C-Na \xrightarrow[CH_{3}Cl]{Excess} CH_{3}-C \equiv C-CH_{3}$$
Compound 'B' is 2-Butyne
Which of the following isomer will have the highest octane number?
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n-Octane
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2-Methylheptane
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2-Methylpentane
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2, 2, 4-Trimethylpentane
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2-Methylhexane
$$IUPAC$$ name of $$CH\equiv C-CH=CH_{2}$$ is
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But-1-yn-3-ene
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But-1-en-3-yne
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But-1-yn-2ene
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None of these
$$\begin{array}{l}{C_2}{H_4} + {H_2} \to {C_2}{H_6},\Delta H = - 32.7KCal\\{C_6}{H_6} + 3{H_2} \to {C_2}{H_{12}},\Delta H = - 49.2KCal\end{array}$$
Resonance energy of Benzene is
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$$-48.9 $$ K Cal
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$$+48.9$$ K Cal
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$$98.8$$ K Cal
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$$-98.8$$ K Cal
Which of the following is general formula of Alkenyne:-
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$$C_{n}H_{2n-2}$$
$$n=4,5,6......$$
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$$C_{n}H_{2n-4}$$
$$n=4,5,6......$$
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$$C_{n}H_{2n-4}$$
$$n=2,3,4......$$
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$$C_{n}H_{2n-2}$$
$$n=2,3......$$
Explanation
Alkenyne is an acyclic hydrocarbon with one $$C-C$$ double bond and one $$C-C$$ triple bond.
E.g. $$CH_{2}=CH-CH_{2}-C\equiv C-CH_{3}$$
General formula: $$C_{n}H_{2n-4},\,\,\,\, n=4,5,6......$$
Predict the correct intermediate and product in the following reaction:
$$H_3C-C \equiv CH\xrightarrow[HgSO_4]{H_2O, H_2SO_2} Intermediate \rightarrow Product $$
$$A$$ $$B$$
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The end product Y of the reaction is:
$$C_2H_5MgBr $$ + $$ S $$ $$ \underrightarrow { \Delta } $$ $$ X $$ $$\underrightarrow { H_3O^+ } $$ $$Y$$
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$$C_2H_6$$
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$$C_2H_5SMgBr$$
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$$C_2H_5SH$$
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$$C_2H_5OH$$
Aqueous solution of potassium acetate is electrolyzed. Possible organic product(s) is/are:
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$$CH_{3}CH_{3}$$
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$$CH_{3}COOCH_{3}$$
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$$CH_{3}CH_{2}CH_{2}CH_{3}$$
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$$Both\ (A)\ and\ (B)$$
A hydrocarbon contains $$7.70\%$$ hydrogen $$0.56\ L$$ of the hydrogen weight $$1.95\ g$$ STP. Then the hydrocarbon is an
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Alkane
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Alkene
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Alkyne
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Arene
In the following compounds, the decreasing order of B.P. is:
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$$(I)>(II)>(III)$$
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$$(I)>(III)>(II)$$
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$$(II)>(III)>(I)$$
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$$(III)>(II)>(I)$$
Which nomenclature is not according to IUPAC system ?
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Which isomer of hexane has only two different sets of structurally equivalent hydrogen atoms ?
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2,2-dimethylbutane
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2-methylpentane
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3-methylpentane
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2,3-dimethylbutane
Alkanes can be prepared from Grignard reagents by reacting with
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Alcohols
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Primary amines
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Alkynes
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All of them
Explanation
Grignard reagents react rapidly with acidic hydrogen atoms in molecules such as alcohols and water to produce alkanes.
A) Grignard reagents react rapidly with acidic hydrogen atoms in molecules such as alcohols and water. When a Grignard reagent reacts with water, a proton replaces the halogen, and the product is an alkane. The Grignard reagent therefore provides a pathway for converting a haloalkane to an alkane in two steps.
B) Though we typically want them to do nucleophilic addition to something, these powerful bases tend to go for protons -- a faster reaction -- if there is just about anything that might make the proton vulnerable. Attaching the proton to an electronegative atom like nitrogen is enough, so primary and even secondary amines will react like that with Grignard reagents. We just deprotonate the amine, the hydrocarbon part of the Grignard reagent ends up an an alkane .
C) We recall that alkynes are more acidic than alkanes. Therefore, the acid–base reaction of an alkyne with a readily available Grignard reagent gives a Grignard reagent of the alkyne. This alkynide ion of the Grignard reagent reacts with carbonyl compounds.
Hence , option D is correct.
Identify the compound which decolourise the brown color of $$Br_{2}$$ dissolved in $$CCI_{4}$$ and gives only two products.
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Which of the following compounds yields only one product on monobromination?
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$$Neopentane$$
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$$Toluene$$
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$$Phenol$$
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$$Aniline$$
Identify the compound which decolorize the brown colour of $$Br_{2}$$ dissolved in $$CCl_{4}$$ and gives only two products?
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0%
0%
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$$2$$
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$$3$$
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$$4$$
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$$5$$
The ratio of hybrid to pure orbitals in benzene molecule is
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3 : 2
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2 : 3
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1 : 2
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4 : 1
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Practice Class 11 Medical Chemistry Quiz Questions and Answers
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