Chemistry · Hydrocarbons · NEET
Look at what the group does to the ring's electrons. If it PUSHES electrons INTO the ring (donates), it is ortho-para directing: -OH, -NH2, -OR, -NHR, -CH3 (and all alkyl), -NHCOCH3. If it PULLS electrons OUT of the ring (withdraws), it is meta directing: -NO2, -COOH, -CHO, -CO-, -CN, -SO3H, -NR3+. One-line test: groups with a lone pair on the atom joined to the ring (N, O) or an alkyl group are ortho-para; groups with a double bond to O/N or a positive charge are meta.
The nitro group pulls electron density away from the ring. Draw the resonance structures: a positive charge (electron-poor spot) appears at the ortho and para carbons. So an electrophile (which wants electrons) avoids ortho and para, because those spots are already electron-poor and their carbocation intermediate would be very unstable. The meta carbon is the least destabilised spot left, so attack happens there. Rule: electron-withdrawing groups make ortho/para electron-poor, so meta is chosen by default.
Halogens do two opposite things. (1) By their -I inductive effect they PULL electrons out, so the ring reacts slower than benzene (deactivating). (2) But halogens also have lone pairs that can donate into the ring by resonance (+M), and this resonance donation reaches the ortho and para carbons, making them a bit richer than meta. So the SPEED is slow (deactivating) but the POSITION is ortho-para. Remember: for halogens, inductive wins for speed, resonance wins for direction.
Both ortho and para form, but usually PARA is the major product for bulky electrophiles or bulky ring groups, because the two ortho positions are crowded next to the existing group (steric hindrance). For small groups, more ortho can form because there are two ortho positions but only one para. For NEET, safest answer: ortho-para directors give a mix, and para is generally major when size matters.
Ortho-para director and activating. Alkyl groups push electrons into the ring by hyperconjugation and a weak +I effect, so they make the ring more reactive and steer the new group to ortho and para. That is why toluene nitrates faster than benzene and gives mainly o- and p-nitrotoluene.
The compound C7H8 (toluene) undergoes: C7H8 --(3Cl2, Δ)--> A --(Br2/Fe)--> B --(Zn/HCl)--> C. The product 'C' is:
Which one of the following compounds does NOT decolourize bromine water?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Donate = ortho/para (and speeds up); Withdraw = meta (and slows down). Groups joined through N or O with a lone pair (-NH2, -OH, -OR) and alkyl groups donate, so ortho/para. Groups with a C=O, C=N, N=O, or a positive charge (-NO2, -COOH, -CHO, -CN, -NR3+) withdraw, so meta. Only exception: halogens are ortho/para directing but deactivating.
No. Most are (like -OH, -NH2, -CH3, -OR). But halogens (-F, -Cl, -Br, -I) are ortho-para DIRECTING yet DEACTIVATING, because their inductive pull (slows reaction) is separate from their resonance donation (chooses ortho-para).
An electron-withdrawing group takes electron density mainly from the ortho and para carbons (you can show this with resonance: positive charge lands on ortho and para). Attack there would give an unstable carbocation. Meta is the position that avoids this instability, so the electrophile goes meta.
-NH2 is a stronger activator than -OH because nitrogen is less electronegative than oxygen, so it donates its lone pair into the ring more easily. Both are strong ortho-para directors and both make the ring react with bromine water without a catalyst.
Yes. The existing group decides WHERE the new group attaches. Ortho-para directors give a mix of ortho and para products (para often major when groups are bulky). Meta directors give mostly the meta product. NEET questions often just ask you to predict this position.
