Optically active $C H_{3} - C H_{2} - O H | C H - C H_{3}$ was found to have lost its optical activity after standing in water containing a few drops of acid, mainly due to the formation of:

C H_{3} - C H_{2} - C H = C H_{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} - C H = C H - C H_{3}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

C H_{3} - C H_{3} | C H - C H_{2} - O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

C H_{3} - C H_{2} - C H_{2} - C H_{2} - O H

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $C H_{3} - C H = C H - C H_{3}$
Optically active $C H_{3} - C H_{2} - O H | C H - C H_{3}$ was found to have lost its optical activity after standing in water containing a few drops of acid, mainly due to the formation of $C H_{3} - C H = C H - C H_{3}$ .
Dehydration takes place to form stable alkene through the formation of stable carbocation. More substituted alkene is more stable.
$C H_{3} - C H_{2} - O H | C H - C H_{3} H^{+} - --- \to - H_{2} O C H_{3} - C H_{2} - C H^{+} - C H_{3} H_{2} O - --- \to - H_{3} O^{+} C H_{3} - C H = C H - C H_{3}$

The alkene thus formed undergoes acid catalysed hydration to regenerated original (but optically inactive) alcohol.
$C H_{3} - C H = C H - C H_{3} H^{+} - - \to H_{2} O C H_{3} - C H_{2} - O H | C H - C H_{3}$

Suggest Corrections

Similar questions

C H_{3} - C H_{2} - O H | C H - C H_{2} - C H_{2} - C | H C H_{3} - C H_{2} - C H_{3}

C H_{3} - C H_{2} - O H | C H - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{2} - C H_{3} | C H - C H_{3}

C H_{3} - C H_{2} - C H_{2} - O H

C H_{3} - C H = C H - O H

C H_{3} - C H = C H - C H_{2} - O H

C H_{3} - C H = C H - O H

Join BYJU'S Learning Program