CH_3-C(CH_3)_2-CH=CH_2 xrightarrow[H_2O_2/OH^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The reaction is Hydroboration-Oxidation. It involves the anti-Markovnikov addition of $H_2O$ across the double bond without rearrangement. $CH_3-C

Vedclass · Accepted Answer

Optically inactive $1^o$ alcohol

Vedclass · Answer

(C) The reaction is Hydroboration-Oxidation. It involves the anti-Markovnikov addition of $H_2O$ across the double bond without rearrangement. $CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow[H_2O_2/OH^{-}]{B_2H_6/THF} CH_3-C(CH_3)_2-CH_2-CH_2OH$ The product is $3,3-	ext{dimethylbutan-1-ol}$,which is a primary $(1^o)$ alcohol and lacks a chiral center,making it optically inactive.

$CH_3-C(CH_3)_2-CH=CH_2 \xrightarrow[H_2O_2/OH^{-}]{B_2H_6/THF} (Z)$ Major product

Similar Questions

The number of sigma $(\sigma)$ bonds in but-$1$-ene is.......

When ethylene bromide is treated with $Zn$,we get

The major product $[B]$ in the following sequence of reactions is
$CH_3-C(CH(CH_3)_2)=CH-CH_2CH_3$ $\xrightarrow[(i) B_2H_6]{(ii) H_2O_2, OH^{-}} [A]$ $\xrightarrow[\Delta]{dil. H_2SO_4} [B]$

Bromination of cyclohexene under the conditions given below yields:
Cyclohexene $\xrightarrow{Br_2/hv}$ ?

Which of the following compounds was the starting material for the oxidation shown below?
$Compound \xrightarrow{KMnO_4 / H^{\oplus}} \text{Product}$

Vedclass Test Series

Exam Paper Generator

Online Exam Module