- ✓deactivates the ring towards electrophilic substitution.
- Bactivates the ring towards electrophilic substitution.
- Crenders the ring basic.
- Ddeactivates the ring towards nucleophilic substitution.
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${N_2}{H_4}\left( l \right) + 2{H_2}{O_2}\left( l \right) \to \mathop {{N_2}\left( g \right) + 4{H_2}O\left( l \right);}\limits_{{\Delta_r} H_1^o = - 818\,kJ/mol} $
${N_2}{H_4}\left( l \right) + {O_2}\left( g \right) \to \mathop {{N_2}\left( g \right) + 2{H_2}O\left( l \right);}\limits_{{\Delta _r}H_2^o = - 622\,kJ/mol} $
${H_2}\left( g \right) + \frac{1}{2}{O_2}\left( g \right) \to \mathop {{H_2}O\left( l \right);}\limits_{{\Delta _r}H_3^o = - 285\,kJ/mol} $
.......$kJ/mol$
Assertion $(A) :$ Treatment of bromine water with propene yields $1-$bromopropan$-2-$ol.
Reason $(R):$ Attack of water on bromonium ion follows Markovnikov rule and results in $1-$bromopropan$-2-$ol.
In the light of the above statements, choose the most appropriate answer from the options given below:
$x\,\,\,\,\,\,\,\,\,\,y\,\,\,\,\,\,\,\,\,\,z$