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CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 5 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Amines
Quiz 5
Among the following compounds, the most basic compounds is:
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Explanation
The most basic compound is [refer image] as it is ready to act as Lewis base by donating electrons and not undergoing resonance as these are no double bonds adjacent to the molecules.
Option D is correct.
Among the isomeric amines select the one with the lowest boiling point.
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Explanation
The boiling point of amine $$\propto$$ tendency to form H-bonds.
No. of H-bonds - $$3^oAmine<2^oAmine<1^oAmine$$[In gas phase]
The $$1^o$$ and $$2^o$$ amine have less steric repulsion than $$3^o$$ amine.
Here in $$3rd$$ compound, it is a tertiary amine that is sterically hindered by the methyl group, that is the reason it has less tendency to form hydrogen bond and has a lowest boiling point.
The answer is C.
In the given pair identify most acidic compound in $$(A)$$ and $$(B)$$. Most basic in $$(C)$$ and $$(D)$$.
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$$A - (I),\ B - (II),\ C - (I),\ D - (II)$$
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$$A - (II),\ B - (I),\ C - (I),\ D - (II)$$
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$$A - (II),\ B - (II),\ C - (II),\ D - (II)$$
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$$A - (I),\ B - (II),\ C - (I),\ D - (I)$$
Explanation
Acidity increases with electron withdrawing group. Ortho effect increases acidity.
Basicity
increases with electron donating group. Ortho effect decreases basicity.
In the given pair of compounds, in which pair second compound has higher boiling point than first compound?
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$$HO - CH_2 - CH_2 - OH$$ and $$CH_3 - CH_2 - OH$$
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Explanation
The extent of hydrogen bonding is more in alcohol than in ether, as alcohol has a hydrogen directly attached to oxygen whereas ether has none. More the alcoholic groups more are the hydrogen bonding.
Now, in the case of amines, Secondary amines have more intermolecular hydrogen bonding than the tertiary amine, due to less steric crowding. More the hydrogen bonding, more the boiling point.
This is because there will be more association between the molecules and more energy will be required to break this association.
Considering the above points last option satisfied the condition.
Thus, it is the correct answer.
In which of the following compound the methylenic hydrogens are the most acidic?
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$$CH_3COCH_2CH_3$$
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$$CH_3CH_2COOC_2H_5$$
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$$CH_3CH_2CH(COOC_2H_5)_2$$
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$$CH_3COCH_2CN$$
Explanation
Here cyanide group pulled the bond pair electron from the chain by the inductive $$(-I$$ effect$$)$$. So, hydrogen of methyl group are because more acidic in nature. $$C-H$$ bond became labile due to pulling of electron by $$CN$$ group.
Correct order of basic strength of given amines is:
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$$\underset{2^o}{Me_2NM} > \underset{1^o}{MeNH_2} > \underset{3^o}{Me_3N > NH_3}$$ (Protic solvent)
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$$\underset{2^o}{Et_2NH} > \underset{3^o}{Et_3H} > \underset{1^o}{EtNH_2} > NH_3$$ (Protic solvent)
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$$Me_3N > Me_2NH > Me - NH_2 > NH_3$$ (Gas phase)
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All are correct
Explanation
Refer to Image
Basic strength (in gas phase)
Because Me is an electron donating group it increases the $$e^-$$ density on N and hence increase its tendency to donate lone pair.
Basicity of amine is directly related to the stability of amine after gaining $$H^+$$.
In the aqueous phase, the substituted ammonium cations gets stabilized not only by e^- releasing effect of the alkyl group (+I) but also by selvation with water molecules. The greater the size of ions, lesser will be the selvation and he less stabilised is the ion. Secondly, when the alkyl group is small like $$-CH_3$$ group, there is no steric hindrance to $$-H$$ bounding.
$$\therefore\quad \begin{matrix} Me_{ 2 }NH \\ 2^o \end{matrix}> \begin{matrix} Me_{ 2 }NH_2 \\1^o \end{matrix}> \begin{matrix} Me_{ 3 }N \\ 3^o \end{matrix}>NH_3$$
$$\therefore \quad \begin{matrix} Et_2NH \\ 2^o \end{matrix}>\begin{matrix} Et_3N \\ 3^o \end{matrix}>\begin{matrix} EtNH_2 \\ 1^o \end{matrix}>NH_3$$
p-aminophenol reacts with one equivalent of acetyl chloride in the presence of pyridine to give mainly.
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Explanation
As $$N$$ is less electronegative than $$0$$, hence nucleophilic attack on acetyl chloride will take place from $$N$$. Hence, acetylation of $$-NH_2$$ takes place.
The amino ketone shown given undergoes a spontaneous cyclization on standing. What is the major product of this intramolecular reaction?
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Which amine yield $$N-$$nitroso amine after treatment with nitrous acid $$(NaNO_2, HCl)?$$
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Explanation
$$\bullet$$ N-Nitroso amine is compound is which group is present.
$$\bullet$$ $$R,{ R }^{ 1 }$$ may be same or different alkyl or aryl group.
Which of the following arylamines will not form a diazonium salt on reaction with sodium nitrite in hydrochloric acid?
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m-Ethylaniline
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p-Aminoacetophenone
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$$4$$-Chloro-$$2$$-nitroaniline
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N-Ethyl-$$2$$-methyaniline
Explanation
$$N-Ethyl-2methyl$$ aniline is a $${ 2 }^{ \circ }$$-secondary aniline (aryl).
But the remaining are primary-aryl-amines which would result in dizonium salt formation.
Which of the following diazonium salt is relatively stable of $$0-5^o$$C?
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$$CH_3CH_2N\equiv N\}^{\oplus}Cl^-$$
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$$CH_3-C(CH_3)_2-N\equiv N\}^{\oplus} Cl^-$$
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$$CH_3-N\equiv N\}^{\oplus}Cl^-$$
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$$(CH_3)_3C-N\equiv N\}^{\oplus}Cl^-$$
Which of following is an example of Pinacol-Diazotization?
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Explanation
Amine groups are converted to $$N_2^+$$ (diazo) groups in presence of $$NaNO_2/HCl$$. When alcohols are adjacent to $$NH_2$$ group Ketones/Aldehydes are formed.
For the diazonium ions the order of reactivity towards diazo-coupling with phenol in the presence of dilute $$NaOH$$ is:
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I $$<$$ IV $$<$$ II $$<$$ III
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I $$<$$ III $$<$$ IV $$<$$ II
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III $$<$$ I $$<$$ II $$<$$ IV
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III $$<$$ I $$<$$ IV $$<$$ II
Explanation
The order of reacting towards diazo coupling with phenol in the presence of dilute $$NaOH$$ of these diazonium ions is :-
Diazotization of n-Bu-$$NH_2$$ with $$NaNO_2/HCl$$ gives _________ isomeric butene.
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$$2$$
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$$3$$
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$$4$$
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$$5$$
Explanation
So total 3 products are possible for diazotisation on n-Butane.
Compare the boiling point of the following two amines.
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Boiling point of $$I > II$$
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Boiling point of $$II > I$$
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Both should have equal boiling points
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It can't be predicted
Which is the strongest base?
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Pyrole
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Aniline
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Pyridine
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Explanation
Hint: A molecule or ion which can easily share its lone pair of an electron
are most basic
Correct Answer:
Option $$D$$
Explanation of Correct Option:
According to the lewis base concept, a stronger base can easily share or donates its lone pair of electrons.
In compound $$D$$, there is one lone pair of electrons present on $$N$$ atom and it can be easily shared with other electrophilic ions or molecules.
Also, it has alkyl groups attached to it which donates by $$+I$$. hence it is the strongest base among all given compounds.
Explanation of incorrect Option:
In the case of compound $$A$$
. the lone pair of nitrogen involved in the aromaticity of the ring, hence cannot be shared to act as a base, therefore it does not act as a strong base.
In the case of compound $$C$$
. The nitrogen center of pyridine features a basic lone pair of electrons. This lone pair does not overlap with the aromatic $$\pi$$ system ring but directly attached with $$sp{^2}$$ hybrid carbon atoms, hence it is less basic than compound $$D$$ where the $$N$$ atom attached with $$sp{^3}$$ hybrid carbon atoms.
In the case of compound $$B$$
. the lone pair of nitrogen is involved in resonance with the ring so the lone pair on nitrogen will not be easily available for donation that is why aniline is a weak base.
Which of the following statement is not true?
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$$CH_3CH_2NH_2$$ is ethanamine
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$$CH_2CH_2OH$$ is an ether
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Cyclopentane and 2-pentene have a molecular formula of $$C_5H_{10}$$
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Alkenes an alkynes are unsaturated
Explanation
$$C{H}_{3}C{H}_{2}N{H}_{2}$$ is Ethanamine or Ethylamine
$$C{H}_{2}C{H}_{2}OH$$ is an alcohol and not an ester.
Cyclopentane and 2-pentene have Molecular formula $${C}_{5}{H}_{10}$$
Alkenes and Alkynes are Unsaturated Compounds.
So, Option B is correct
Which of the following diazonium salt is relatively stable at $$0-{ 5^0 }^{ }C$$?
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$${ CH }_{ 3 }-N\equiv N{ \} }^{ \oplus }C{ l }^{ - }$$
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$${ \left( { CH }_{ 3 } \right) }_{ 2 }CH-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$$
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$${ C }_{ 6 }{ H }_{ 5 }-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$$
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$${ \left( { CH }_{ 3 } \right) }_{ 3 }C-N\equiv N{ \} }^{ \oplus }{ Cl }^{ - }$$
Explanation
The $$+I$$ effect which stabilizes the $${ N }^{ + }$$ion is in the following order $${ C }_{ 6 }{ H }_{ 5 }-<{ CH }_{ 3 }-<{ \left( { CH }_{ 3 } \right) }_{ 2 }-<{ \left( { CH }_{ 3 } \right) }_{ 3 }$$
$$\therefore $$ The $${ \left\{ { \left( { CH }_{ 3 } \right) }_{ 3 }C-N\equiv N \right\} }^{ + }{ Cl }^{ - }$$ is most stable
What product is formed on heating two molecules of urea?
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Uric Acid
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Biuret
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Nitrogen
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Carbon dioxide
Explanation
Urea is $$NH_2-\overset{\overset{O}{||}}{C}-NH_2$$
When $$2$$ molecules of urea is heated, biuret is formed.
$$2NH_2-\overset{\overset{O}{||}}{C}-NH_2 \longrightarrow NH_2-\overset{\overset{O}{||}}{C}-NH-\overset{\overset{O}{||}}{C}-NH_2$$
Biuret
The most basic among the following is:
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$${ H }_{ 3 }C-{ NH }_{ 2 }$$
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$${ CH }_{ 3 }-{ CH }_{ 2 }-NH_2$$
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$${ H }_{ 3 }C-NH-{ CH }_{ 3 }$$
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$${ H }_{ 3 }C-{ CH }(CH_3)-NH_2$$
Explanation
$${ H }_{ 3 }-C$$$$\rightarrow -NH\leftarrow -C-{ H }_{ 3 }$$
$$+I$$ effects of methyl group
Here, also methyl group attached with the N-atom. The inductive effect of methyl group increases the density of electron on N-atom. It becomes more basic.
$$H-COOH > Ph-COOH > CH_{3}-COOH$$ (acidic nature)$$CF_{3}-COOH > CCI_{3}-COOH > CH_{2}-COOH$$ (Acidic nature)$$N- methylmethanamine > methanamine > N, N- dimethylmethanamine $$ (Basic nature in aqueous phase)$$N,N- diethlethanamine > N-ethylethanamine > Ethanamine $$ (Basic nature in aqueous phase)
Then the correct order is:
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$$i,ii,iv$$
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$$ii,iii,iv$$
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$$i,ii,iii$$
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$$i,ii,iii,iv$$
Explanation
In aqueous phase, the basicity of amines depends upon:
$$(i)$$ Inductive effect: which fovours tertiary amine to be more basic due to greater $$+I$$ effect of $$3$$alkyl group directly attached to nitrogen.
$$(ii)$$ Solution Effect: Which favours primary amine to be more basic due to greater solution of its cation.
$$(iii)$$ Steric Factor: Which cause tertiary amine to be less basic due to steric hinderance .
What is the major organic product of the following reaction?
$$ { CH }_{ 3 }{ CH }_{ 2 }{ NHCH }_{ 2 }{ CH }_{ 3 }\longrightarrow $$?
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Which is most basic in aqueous solution?
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$${ CH }_{ 3 }-{ NH }_{ 2 }$$
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$${ (CH }_{ 3 }{ ) }_{ 2 }{ NH }$$
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$${ (CH }_{ 3 }{ ) }_{ 3 }N$$
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$$ph-{ NH }_{ 2 }$$
Explanation
$${ CH }_{ 3 }\rightarrow \begin{matrix} N \\ \uparrow \\ { CH }_{ 3 } \end{matrix}\leftarrow { CH }_{ 3 }$$
$$+I $$ effects of Three methyl group
Here, it is 3 amine and three methyl group attached with N-atom. The inductive effect of three methyl group and lypez conjugation effect of $$9-\alpha -H$$ atom, So this is more stable and basic in nature.
Which of the following is most basic ?
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Explanation
Here, lone pair of electrons are available on the $$N-$$ atom and it can easily donate electrons and $$pK_b$$ value is also low.
Select the basic strength order of following molecules.
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$$III > II > I$$
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$$II > III > I$$
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$$I > III > II$$
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$$III > I > II$$
Explanation
Since in (III) lone pair is only involved in delocalisation, in (II) nitrogen is attached to the carbonyl (withdrawing group) and one alkyne group (donating group+I effect) which have the more basic character them the (I) in which (N) is
attached to the two group i.e benzene and carbonyl.
Which of the following conditions is most suitable for the above reaction?
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strongly acidic
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strongly alkaline
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Slightly alkaine
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neutral
Explanation
In this reaction, primary aromatic amine is treated with nitrous acid ($${ HNO }_{ 2 }$$).
$${ HNO }_{ 2 }$$ can be obtained by dissolving solid
$${ NaNO }_{ 2 }$$ in aqueous acidic solution.
Which of the following will accept $$H^{+}$$ from $$NH_{4}^{+}$$ ion?
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$$CH_{3}-CH_{2}-NH_{2}$$
Explanation
Since the aromatic amines (aniline) are less basic than aliphatic amines.
Because lone pair of nitrogen in aromatic amines are involved in resonance.
Decreasing order of basicity of the three isomers of nitro aniline is:
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$$p-nitroaniline> o-nitroaniline> m-nitroaniline$$
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$$p-nitroaniline> m-nitroaniline> o-nitroaniline$$
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$$m-nitroaniline> p-nitroaniline> o-nitroaniline$$
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$$m-nitroaniline> o-nitroaniline> p-nitroaniline$$
Explanation
Nitro group is an electron-withdrawing group. It mainly withdraws electrons from the ortho and para positions. Therefore para isomer is less basic than meta isomer.
Ortho substituted anilines are generally weaker bases than aniline irrespective of the electron releasing or electron-withdrawing nature of the substituent.
Hence decreasing order of basicity of the isomers of nitroaniline:
$$m-nitroaniline>p-nitroaniline>o-nitroaniline$$
The order of basic strength of the given basic nitrogen atoms is :
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$$III > II > I > IV$$
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$$III > I > II > IV$$
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$$I > III > II > IV$$
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$$II > III > I > IV$$
Explanation
Since $$(II)$$ is more basic then $$(III)$$ due $$(II)$$ nitrogen get negative charge due resonance hence it is more basic.
There for correct order is $$II$$>$$III$$>$$I$$>$$IV$$
D)
The correct sequence regarding base strength of aliphatic amines in aqueous solution is:
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$${R}_{3}N> {R}_{2}NH> R{NH}_{2}> {NH}_{3}$$
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$${R}_{2}NH> R{NH}_{2}> {R}_{3}N> {NH}_{3}$$
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$${R}_{2}NH> {R}_{3}N> R{NH}_{2}> {NH}_{3}$$
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$$R{NH}_{2}> {R}_{2}NH> {R}_{3}N> {NH}_{3}$$
Explanation
Since the bascities of amine depend upon the no of alkyl group directly attached to the nitrogen directly due to $$+I$$ effect to alkyl group increased the bascities of amine.
The increasing order of basicity of the given compounds is:
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(b)< (a)< (c)< (d)
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(b)< (a)< (d)< (c)
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(d)< (b)< (a)< (c)
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(a)< (b)< (c)< (d)
Explanation
Order of base nature depends on electron donation tendency.
In compound (b) nitrogen is $$sp^{ 2 }$$ hybridized so least basic among all given compound
Compound (C) is a very strong nitrogeneous organic base as a lone pair of one nitrogen delocalize in resonance and make another nitrogen negatively charged and conjugate acid have two equivalent resonating structure.
Thus it is most basic in given compound.
Compound (d) is secondary amine more basic than compound (a) which is a primary amine.
Therefore, increasing order of basicity is
(b) < (a) < (d) < (c)
The correct order of basic strength of the compounds?
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$$I > II > III$$
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$$II > I > III$$
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$$III > II > I$$
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$$II > III > I$$
Explanation
the strength of basic nature of compounds depends on the ability to easily donate the lone pair of electrons.
IN $$(i)$$ lone pair is easily approachable so it has more basicity . in $$(ii)$$ the lone pair is in different plane. while in $$(iii)$$ the lone pair is in conjugation . hence order of basicity is $$I>II>III$$
By heating ammonium chloride with two equivalents of formaldehyde it forms:
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dimethylamine
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ethylamine
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methylamine
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ammonium formate
Explanation
$$6HCHO + 4N{H}_{4}Cl \rightarrow ({C{H}_{2}}_{6}){N}_{4} + 4HCl + 6{H}_{2}O$$
So Methylamine is formed
Which of the following Statements is/are correct?
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$$(SiH_3)_3N $$ is weaker base than $$(CH_3)_3N$$.
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Amongst $$Cl-Cl$$, $$N-F$$,$$C-F$$ And $$O-F$$ bond is most polar
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$$OF_2$$ has higher dipole moment than $$H_2O$$
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An alchohol is less volatile than other having same molecular formula on account of intermolecular hydrogen bonding
Explanation
$$A.$$ $${(SiH_3)}_3N$$ is a weaker base than $${(CH_3)}_3N$$.
In $${(CH_3)}_3N$$ the lone pair is concentrated on the $$N-atom$$. But in case of Trisillyl amine, the lone pair on nitrogen gets delocalised on the $$3$$ silicon atoms bonded to it.
Since, base strength is a measure of tendency to donate lone pair, trimethylamine is stranger base.
$$D.$$ In ethers there is no intermolecular hydrogen bond, where as in alcohol, intermolecular hydrogen bond is present which makes them less volatile and have high boiling point.
$${C}_{4}{H}_{11}N(X)+H{NO}_{2} \rightarrow {C}_{4}{H}_{10}O$$($${3}^{o}$$ alcohol)
Hence, $$X$$ will not give:
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carbylamine reaction
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hoffmann mustard oil reaction
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diazorium salt (as the intermediate)
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hofmann's bromamide reaction
Explanation
$$X$$ does not undergo carbylamine reaction because this reaction requires only primary amine. In the above question the amine is secondary amine because $$X$$ is attached to the amine.
$$sp^3-$$hybridised nitrogen is present in:
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$$CH_2 = CH - \overset{\oplus}{N}H_3$$
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Explanation
A) $$Sp^{2}$$ hybridized Nitrogen
B) $$S p^{2}$$ hybridized Nitrogen due to participation
of lone pair into ring.
c) $$C H_{2}=C H-NH_3+ \quad\left(S p^{3}\right.$$ hybridised nitrogen $$)$$
D) (sp $$^{2}$$ hybridised) as nitrogen is in conjugation with double bond.
So, option C) is correct.
The product $$B$$ of the following sequence, would be?
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$$I$$
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$$II$$
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$$III$$
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$$IV$$ and $$V$$
The correct order of increasing basic nature for the bases $$NH_3,CH_3NH_2$$ and $$(CH_3)_2NH $$ is?
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$$(CH_3)_2NH < NH_3 < CH_3NH_2$$
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$$NH_3 < CH_3 NH_2 <(CH3)_2NH $$
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$$CH_3NH_2 <(CH_3)_2NH < NH_3$$
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$$CH_3NH_2 < NH_3 < (CH_3)_2NH$$
Explanation
$$(H_3C)_2NH$$ , $$CH_3NH_2$$ , $$NH_3$$
Among the above molecules $$(CH_3)_2 NH$$ is most basic as two electron donating group are attached to it which increase its basicity is
$$NH_3 < CH_3NH_2 < (CH_3)_2NH$$
The end product (Z) of the following reaction is?
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A cyanide
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A carboxylic acis
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An amine
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An arene
Find product in the following reaction is Y. What is Y?
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benzamide
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benzophenone
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benzoic acid
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benzaldehyde
Explanation
In the given reaction, reduction of the cyanide group takes place followed by steam distillation to form benzaldehyde as a product.
$$CH_3Cl \xrightarrow{AgCN} A$$ (major) $$\xrightarrow{H-2O^+} C + D$$ here '$$C$$' is formic acid '$$D$$ is:'
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$$2^o$$ amine
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Carboxylic acid
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Alkyl Cyanide
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$$1^o$$ amine
Explanation
$$\text { Hence correct Answer - D }$$
Reductive amination of $$A$$ forms:
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The correct order of boiling points of the following amines $${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 },\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH,{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$ is :
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$${ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$
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$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$
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$${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2} >\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$
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$$\left( { C }_{ 2 }{ H }_{ 5 } \right) _{ 2 }NH>{ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }>{ C }_{ 2 }{ H }_{ 5 }N\left( { CH }_{ 3 } \right) _{ 2 }$$
Explanation
Because $${ C }_{ 4 }{ H }_{ 9 }{ NH }_{ 2 }$$ is primary amine and can form hydrogen bonding more than secondary amine $$({ { C }_{ 2 }{ H }_{ 5 } })_{ 2 }NH$$ and tertiary amine $${ C }_{ 2 }{ H }_{ 5 }N{ ({ CH }_{ 3 }) }_{ 2 }$$ .More hydrogen bonding leads to strong bonding between the molecules, hence increases the boiling point.
Which of the following is the least basic?
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All are equally basic
Explanation
Since $$NO_{2}$$ is strong withdrawing group than $$OCH_{3}$$ and $$C_{6}H_{5}$$ , hence
withdraw more electron density from benzene and leads to lone pair of $$NH_{2}$$ to be dicolocalised at faster rate.
In the above reaction sequence (D) is?
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Suitable explanation for the order of basic character, $$(CH_{3})_{3}N < (CH_{3})_{2}NH$$, is:
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steric hindrance by bulky methyl group
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higher volatility of $$3^{\circ}$$ amine
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stability of conjugate acid due to solvation
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all of these
Explanation
$${ ({ CH }_{ 3 }) }_{ 3 }$$ N is less basic due to steric hindrance by bulky methyl group prevent any acid to take a lone pair of nitrogen.
With which one of the following substrate do diazonium salt reacts to form a colourful compound?
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Explanation
The answer is refer to image.
In the reaction
$$Me_{3}N \xrightarrow{BrCN} A \xrightarrow[ii)\Delta]{i)H_{2}O} B$$.
B is
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$$MeNH_{2}$$
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$$Me_{3}NO$$
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$$Me_{2}NH$$
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$$Me_{2}NCOOH$$
Explanation
$$Me_3N\xrightarrow {BrCN}{}\underset{A}{Me_2N}-CN$$
$$Me_2N-Cn\xrightarrow[\triangle ]{H_2O}Me_2NCOOH$$
The reagents/conditions K, L and M are respectively:
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$$Ba(OH)_{2}$$, $$KCN$$ and $$CHCl_{3} / NaOH$$ heat, $$H_{3}O^{+}$$
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$$H_{2}O$$ (boil), $$CO_{2} / KOH$$ and $$Ac_{2}O - AcONa$$, heat
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Steam ; $$CHCl_{3} / NaOH$$, heat, $$H_{3}O^{+}$$ and $$AC_{2}O - AcONa$$, heat
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$$Cu_{2}Cl_{2} / HCl$$, $$NaOH$$ / high P and $$CO_{2}|NaOH$$, heat
Explanation
$$ \text { Honre option } A \text { is correct } $$
Which of the following compounds is an imine?
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Explanation
The structure of imine is the given figure.
The nitrogen atom can be attached to hydrogen or an organic group.
Hence compound $$B$$ is an imine.
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Practice Class 12 Engineering Chemistry Quiz Questions and Answers
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