NEET Questions Solved

Which is the most suitable reagent among the following to distinguish compound (III) from rest of the compounds?

I. CH₃-C $\equiv$ C-CH₃

II. CH₃-CH₂-CH₂-CH₃

III. CH₃-CH₂-C $\equiv$ CH

IV. CH₃-CH=CH₂

(a) Br₂/CCl₄ (b) Br₂/CH₃COOH

(c)Alk.KMnO₄ (d) Ammonical AgNO₃

#Alkanes, Alkenes and Alkynes - Chemical Properties

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(d) It may be noted that only the terminal alkynes react with ammonical silver nitrate. Therefore, this reaction can be used to distinguish between 1-alkynes and others such as alkane, alkenes and non-terminal alkynes.

${CH}_{3} - {CH}_{2} - C \equiv CH + 2 [Ag ({NH}_{3})_{2}] {NO}_{3} \to {CH}_{3} {CH}_{2} C \equiv CAg ↓$

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Difficulty Level:

(a)
20%
(b)
14%
(c)
24%
(d)
43%

Which of the following reagents will be able to distinguish between 1-butyne and 2-butyne?

(a) NaNH₂ (b) HCl (c) O₂ (d) Br₂

#Alkanes, Alkenes and Alkynes - Chemical Properties

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(a) NaNH₂ is used to distinguish between 1-butyne and 2-butyne.

CH₃CH₂C $\equiv$ CH $\overset{{NaNH}_{2}}{\to}$ CH₃CH₂C $\equiv$ CNa + 1/2 H₂ $↑$

CH₃-C $\equiv$ C-H $\overset{{NaNH}_{2}}{\to}$ No reaction

Difficulty Level:

(a)
62%
(b)
14%
(c)
8%
(d)
17%

A compound is treated with NaNH₂ to give sodium salt. Identify the compound.

(a) C₂H₂ (b) C₆H₆

#Alkanes, Alkenes and Alkynes - Chemical Properties

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(a) Sodamide is strong base, attract the more acidic hydrogen and gives sodium salt

H-C $\equiv$ C-H + NaNH₂ $\to$ H-C $\equiv$ C^-Na⁺ + 1/2 H₂

Sodium ethynide

Difficulty Level:

(a)
63%
(b)
14%
(c)
11%
(d)
14%

When chlorine is passed through propene at 400 $°$ C, which of the following is formed?

(a) PVC (b) Allyl chloride

#Alkanes, Alkenes and Alkynes - Chemical Properties

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(b) When chlorine gas is reacted with propene at high temperature (400 $°$ C), then substitution takes place in place of addition reaction. Hence, allyl chloride is formed.

CH₃-CH=CH₂ + Cl₂ $\overset{400 ° C}{\to} \underset{|}{C} H_{2} - CH = {CH}_{2} + HCl Cl$

Difficulty Level:

(a)
28%
(b)
47%
(c)
5%
(d)
22%

When acetylene is passed through $dil . H_{2} {SO}_{4}$ in presence of ${HgSO}_{4}$ , the compound formed is

(a) ether (b) ketone

#Alkanes, Alkenes and Alkynes - Chemical Properties

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(d) Hydration of alkyne,

$CH \equiv CH$ + $H_{2} O \overset{{Hg}^{2 +} / H^{+}}{\to}$ $\overset{Tautomerisation}{\to}$ ${CH}_{3} - CHO$

Difficulty Level:

(a)
8%
(b)
21%
(c)
14%
(d)
59%

CH₃-C $\equiv$ C-H $\overset{{CH}_{3} MgBr}{\to}$ CH₄ + (A) $\to_{(ii) H_{2} O / H^{+}}^{(i) {CO}_{2}}$ (B); (B) will be

(1) H₃C-CH₂-CH₂-COOH (2) CH₃-C $\equiv$ C-CH₃

(3) H₃C-C $\equiv$ C-COOH (4) H₃C-CH=CH-COOH

#Alkanes, Alkenes and Alkynes - Chemical Properties

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(3) CH₃-C $\equiv$ C-H $\overset{{CH}_{3} MgBr}{\to}$ H3C-C $\equiv$ C^-Mg⁺Br $\to_{(ii) H_{2} O / H^{+}}^{(i) {CO}_{2}}$ H₃C-C $\equiv$ C-COOH

(A) (B)

Difficulty Level:

(a)
19%
(b)
12%
(c)
48%
(d)
23%

The monochlorinated products (excluding stereo-isomers) obtained from the reaction is:

(1) 4 (2) 5 (3) 6 (4) 7

#Alkanes, Alkenes and Alkynes - Chemical Properties

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Every unique position with replaceable H will result in a monochlorinated product

(1,5) (2,4 and (6,8) are identical positions. Therefore, five monochlorinated products can be formed.

Difficulty Level:

(a)
15%
(b)
66%
(c)
14%
(d)
7%