Write the product $(C H_{3})_{3} C - O H C u - -- \to 573 K$

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Solution

Tertiary alcohol when react with copper at 573K dehydration takes place and alkene is formed while primary and secondary alcohol on reaction gives aldehyde and ketone by dehydrogenation.

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Similar questions

Give reasons for the following:

(i) o-nitrophenol is more acidic than o-methoxyphenol.

(ii) Butan-1-ol has a higher boiling point than diethyl ether.

(iii) $(C H_{3})_{3} C - O - C H_{3}$ on reaction with HI gives $(C H_{3})_{3} C I -$ and $C H_{3} - O H$ as the main products and not $(C H_{3})_{3}$ C–OH and $C H_{3} - I$ .

Q. The reaction

(C H_{3})_{3} C - B r H_{2} O ⟶ (C H_{3})_{3} - C - O H

is a/an :

Q. Give reasons of the following.
(i) Phenol is more acidic than methanol.
(ii) The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle

(109^{\circ} 28^{^{'}})

.
(iii)

(C H_{3})_{3} C - O - C H_{3}

on reaction with HI gives

(C H_{3})_{3} C - I a n d C H_{3} - O H

as the main products and not

(C H_{3})_{3} C - O H a n d C H_{3} - I

Q. Assertion :Statement-1 :

(C H_{3})_{3} S i - O H

is more acidic than

(C H_{3})_{3} C - O H

. Reason: Statement-2 :

(C H_{3})_{3} S i - O H

has back bonding.

(C H_{3})_{3} C - O H

on treatmant with

N a C l

in aqueous medium gives:

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Write the product (CH3)3C−OHCu−−−→573K

Tertiary alcohol when react with copper at 573K dehydration takes place and alkene is formed while primary and secondary alcohol on reaction gives aldehyde and ketone by dehydrogenation.

Write the product $(C H_{3})_{3} C - O H C u - -- \to 573 K$