Hydrocarbons - Result Question 3
4. The major product of the following reaction is
$\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { (ii) DI }]{\text { (i) } \mathrm{DCl} \text { (1 equiv.) }}$
(2019 Main, 9 Aprill)
(a) $\mathrm{CH}_3 \mathrm{CD}(\mathrm{Cl}) \mathrm{CHD}(\mathrm{I})$
(b) $\mathrm{CH}_3 \mathrm{CD}_2 \mathrm{CH}(\mathrm{Cl})(\mathrm{I})$
(c) $\mathrm{CH}_3 \mathrm{CD}(\mathrm{I}) \mathrm{CHD}(\mathrm{Cl})$
(d) $\mathrm{CH}_3 \mathrm{C}(\mathrm{I})(\mathrm{Cl}) \mathrm{CHD}_2$
Show Answer
Answer:
Correct Answer: 4. (d)
Solution:
- The major product obtained in the given reaction is $\mathrm{CH}_3 \mathrm{C}$ (I) (Cl) $\mathrm{CHD}_2$.
$\mathrm{CH}_3 \mathrm{C} \equiv \mathrm{CH} \xrightarrow[\text { 2. DI }]{\text { 1. } \mathrm{DCl} \text { (1equiv.) }} \mathrm{CH}_3 \mathrm{C}(\mathrm{I})(\mathrm{Cl}) \mathrm{CHD}_2$
Addition in unsymmetrical alkynes takes place according to Markovnikov’s rule. Reaction proceed as follows :
BETA
