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CBSE Questions for Class 12 Engineering Chemistry Amines Quiz 15 - MCQExams.com
CBSE
Class 12 Engineering Chemistry
Amines
Quiz 15
The correct order of the basic strength of following compounds is:
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$$4>2>3>1$$
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$$4>2>1>3$$
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$$3>1>2>4$$
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$$1>3>4>2$$
Explanation
$$\text{Higher the electron density on electron-donating group higher is the basic strength.}$$
$$\text{In compound 3 the electron density is highest because of }+ I\text{ effect of methyl groups attached to N}$$. $$\text{Hence it has the highest basicity.}$$
$$\text{In compound 1 there are two amine groups attached to the carbon in which one of the amine group is always}$$$$\text{available for donation. Hence it has the second-highest basicity.}$$
$$\text{In compound 2 again the ethyl group increases the basic strength of amine due to its }+ I\text{ effect but lower}$$ $$\text{than that of as in compound 3.}$$
$$\text{The compound 4 has the least basic strength because the carbonyl group attached to amine decreases the }$$$$\text{basicity by withdrawing the lone pair from Nitrogen through resonance effect.}$$
$$\text{The correct order of basic strength is - }3>1>2>4$$
Which of the following is the correct order of basic character ?
I. 1-Amino Propane
II. Ethanamide
III. Guanidine $$[HN=C(NH_2)_2]$$
IV. Aniline
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I > II > III > IV
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III > I > IV > II
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IV > III > I > II
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III > II > I > IV
In the Gabriel's phthalimide synthesis, phthalimide is treated first with
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$$C_2H_5I / KOH$$
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Ethanolic Na
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Ethanol and $$H_2SO_4$$
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Ether and $$LiAlH_4$$
Action of diazomethane on phenol liberates
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$$ O_2 $$
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$$ H_2 $$
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$$ N_2 $$
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$$ CO_2 $$
Explanation
Option C is correct.
When aniline is treated with sodium nitrate and hydrocholric acid at O.C, it gives
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Phenol and $$ N_{2} $$
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Diazonium salt
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Hydrazo compound
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No reaction take place
Explanation
$$C_{6}H_{5}NH_{2} \xrightarrow {NaNO_{2} + HCl\,O^{\circ}C} C_{6}H_{5}N_{2}Cl$$
Option B is correct.
The correct order of decreasing basicity of the following
compounds is:
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$$I > II > III > IV$$
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$$II > I > III > IV$$
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$$III > IV > I > II$$
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$$II > I > IV > III$$
Explanation
The order of decreasing basicity of the given compounds is (C) $$III > IV > I > II.$$
As the $$−NH_2$$ group has a lone pair of electrons on nitrogen to donate to electron-deficient species.
Hence, its power to donate electrons is increased when an electron donation group $$(+R, +I)$$ is attached and its donating power is decreased when an electron-withdrawing group is attached at its para position.
Since $$NO_2$$ is an electron-withdrawing group hence it decreases the electron density over $$−NH_2$$ group hence it decreases the basicity.
On the other hand $$−OCH_3 (+M \text{ effect})$$ group increases the electron density more than the $$CH_3\ group (+I \text{effect}).$$
The diazonium salts are the reaction product in presence of excess of mineral acid with nitrous acid and
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Primary aliphatic amine
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Secondary aromatic amine
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Primary aromatic amine
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Tertiary alipathic amine
Explanation
The diazonium salts are the reaction product in presence of excess of mineral acid with nitrous acid and primary aromatic amine
Hence C is correct answer
$$(RCO)_2 NH$$ is
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Primary amine
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Secondary amine
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Secondary amide
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Tertiary amide
Explanation
$$R-\overset {\overset {\displaystyle O}{||}}{C}-NH_2 \Rightarrow \text {Primary amide}$$
$$(R-\overset {\overset {\displaystyle O}{||}}{C})_2-NH \Rightarrow \text {Secondary amide}$$
Option C is correct.
$$ C_{3}H_{9}N $$ represents
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Primary amine
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Secondary amine
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Tertiary amine
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All of these
Explanation
$$C_{3}H_{9}N$$ can form all the 3 amines.
$$CH_{3}\underset{1^{o} amine}{CH_{2}CH_{2}}-NH_{2},CH_{3}-\underset{2^{o} amine}{CH_{2}-NH}-CH_{3}$$
Rank the following compound in the order of decreasing basicity.
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$$I, \ II,\ III$$
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$$III, \ II, \ I$$
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$$II, \ I, \ III$$
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$$III, \ I, \ II$$
Explanation
$$III$$ is the stronger base due to lack of delocalisation of lone pair of electron. $$II$$ is weaker base than $$I$$ due to electron withdrawing resonance effect of $$-NO_2$$.
In this reaction $$ C_{6}H_5{}NH_{2}+HCl+NaNO_{2}\rightarrow X $$
product X is
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Aniline hydrochloride
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Nitro aniline
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Benzenediazonium chloride
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None of these
Explanation
$$C_{6}H_{5}NH_{2} + HCl + NaNO_{2} \rightarrow C_{6} H_{5} N_{2} Cl$$
Benzenediazonium chloride
Option C is correct.
Urotropine is
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Hexamethylene tetramine
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Hexaethylene tetramine
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Hexamethylene diamine
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None of these
Explanation
$$\text{Urotropin is hexa-methylene tetramine .It is used as a medicine to treat urinary infections.}$$
Which of the following statements are true?
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$$(I)$$ and $$(III)$$ are modest Bronsted bases, whereas $$(II)$$ is not.
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In $$(III),\ N^a$$ in more basic than $$N^b$$
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When $$(III)$$ is protonated in the presence of a strong acid, protonation occurs at $$C-2$$
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$$(I)$$ and $$(II)$$ compounds are aromatic.
Explanation
But (II) is not protonated. Hence statement (A) is true
Hence, statement (B) is wrong
Statement (C) is true
Statement (D) : As per Huckels rule both I and II are aromatic.
Which of the following statements are correct?
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$$(I)$$ and $$(II)$$ are aromatic and have equal basic strength
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$$(I)$$ is aromatic $$(II)$$ is anti-aromatic, but $$(II)$$ is a stronger base than $$(I)$$
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The basicity order of above compounds is $$(IV) > (III) > (II) > (I)$$
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The conjugate acid of $$(IV)$$ is more stabilised than the conjugate acid of $$(II)$$
Explanation
In (IV), the lone pair $$e, s$$ of two $$N$$ makes it more basic and does not delocalises in the benzene ring.
In (III) no delocalisation of $$NH_2$$. $$L.P$$ of $$e, s$$ ($$sp^3$$)
In (II) no delocalisation of $$LP$$ of $$e, $$ on $$N$$ ($$sp^2$$)
In (I), delocalisation of $$L.P$$ of $$e, s$$ on $$N$$, via resonance
II is aromatic compound.
Hence, the order of basic character is $$(IV)>(III)>(II)>(I).$$
Identify the product in following order
$$ 3,4,5-Tribromoaniline\xrightarrow[(ii)H_{3}PO_{2}]{(i)diazotizaton}? $$
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3,4,5- Tribromobenzene
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1,2,3-- Tribromobenzene
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2,4,6-- Tribromobenzene
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3,4,5-- Tribromo nitro benzene
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3,4,5-- Tribromo phenol
Which statement is not correct ?
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Amines from hydrogen bond
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Ethyl amine has higher boiling point than propane
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Methyl amine is more basic than ammonia
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Diamethyl amine is less basic than methyl amine
Explanation
In methyl amine only one electron releasing group is present but in dimethyl amine two electron releasing groups are present which increase the basicity higher in dienethyl amine.
Which of the following amines can be prepared by Gabriel synthesis.
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Isobutyl amine
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$$2-$$ Phenylethylamine
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$$n-$$methylanime
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Anilline
Explanation
Gabriel synthesis is used for the preparation of primary amines. Phthalimide on treatment with ethanolic potassium hydroxide forms potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis produces the corresponding primary amine.
Nitrobenzene on reduction with $$Zn/NaOH$$ gives:
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$${C}_{6}{H}_{5}-N=N-{C}_{6}{H}_{5}$$
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$${C}_{6}{H}_{5}-NH-NH-{C}_{6}{H}_{5}$$
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$${C}_{6}{H}_{5}NHOH$$
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$$C_6 H_5 - \underset{O}{\underset{\downarrow}{N}} = N - C_6H_5$$
Explanation
Nitrobenzene on reduction with $$Zn/NaOH$$ gives Azobenzene.
Which of the following is most basic?
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$$CH_3NH_2$$
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$$(CH_3)_2NH$$
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$$(CH_3)_3N$$
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$$C_6H_5NH_2$$
$$C_3H_9N$$ does not show
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Primary amine
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Quaternary salt
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Tertiary amine
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Secondary amine
The major product of the reaction is
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Explanation
Reaction taking place is
So, the product formed is $$\mathrm{CH_3-CH(CH_3)-CH_2-CH(OH)-COOH}$$.
Hence, Option "C" is the correct answer.
The correct increasing order of basic strength for the following compounds
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$$I>III>II$$
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$$III>II>I$$
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$$II>I>III$$
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$$III>I>II$$
Explanation
Electron donating $$-CH_3$$ group increases the basicity while electron withdrawing $$-NO_2$$ group decreases the basicity of amines.
Hence correct option is (D) : III > I > II
$$C_3N_9N$$ cannot represent
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$$1^o$$ amine
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$$2^o$$ amine
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$$3^o$$ amine
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quaternary ammonium salt
Explanation
Quaternary ammonium salt, as it carries a positive charge over nitrogen e.g.$$(CH_3)_4N^+Cl^-$$
Hence option (D) is correct.
Which of the following has highest $$pK_{b}$$ value?
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$$(CH_{3})CNH_{2}$$
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$$NH_{3}$$
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$$(CH_{3})_{2}NH$$
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$$CH_{3}NH_{2}$$
Explanation
Aniline gets basicity, lower is the $$pK_{b}$$ value. Since $$NH_{3}$$ is the weakest base, hence it has highest $$pK_{b}$$ value.
Hence option (B) is correct.
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