MCQ
The decrease in atomic volume from $Cr$ to $Cu$ is very negligible because
- AIncrease in nuclear change
- ✓Screening effect
- CUnpaired electrons of $Cr$
- DNone
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$\begin{array}{*{20}{c}}
{C{H_3} - C = C{H_2}C{H_2}OH} \\
{|\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\
{\,C{H_3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}
\end{array}$ is
$A+B\xrightarrow{K}C$
Rate Order
| Column $I$ | Column $II$ |
| $(A)$ $\mathrm{CH}_3-\mathrm{CHBr}-\mathrm{CD}_3$ on treatment with alc. $\mathrm{KOH}$ gives $\mathrm{CH}_2=\mathrm{CH}-\mathrm{CD}_3$ as a major product. | $(P)$ $E1$ reaction |
| $(B)$ $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CH}_3$ reacts faster than $\mathrm{Ph}-\mathrm{CHBr}-\mathrm{CD}_3$. | $(Q)$ $E2$ reaction |
| $(C)$ $\mathrm{Ph}-\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{Br}$ on treatment with $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OD} / \mathrm{C}_2 \mathrm{H}_5 \mathrm{O}^{-}$ gives $\mathrm{Ph}-\mathrm{CD}=\mathrm{CH}_2$ as the major product. | $(R)$ $E1$ cb reaction |
| $(D)$ $\mathrm{PhCH}_2 \mathrm{CH}_2 \mathrm{Br}$ and $\mathrm{PhCD}_2 \mathrm{CH}_2 \mathrm{Br}$ react with same rate. | $(S)$ First order reaction |