MCQ
Comment on stereochemistry of products


- ✓diastereomers
- Bracemic
- Csingle stereoisomer
- Dmeso

Two chiral carbons in the product. Configuration of one carbon is fixed. Attack on the carbonyl carbon can be from two sides. Hence products obtained will be diastereomers.
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$\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{ r }{ a _0}\right) e ^{- r / 2 a _0}$
At $r=r_0$, radial node is formed. Thus, $r_0$ in terms of $a_0$
$(II)$ $1,3-$dihydroxy benzene
$(III)$ $1,4-$dihydroxy benzene
$(IV)$ Hydroxy benzene
The increasing order of boiling points of above mentioned alcohols is

