Secondary haloalkanes (CH₃CHClCH₃) react slower than primary haloalkanes (CH₃CH₂Br; CH₃CH₂Cl), as secondary tend to react via both the S_N1 and S_N2 mechanisms rather slowly. Primary haloalkanes react faster than secondary; they react via S_N2. Tertiary haloalkanes ((CH₃)₃CCl) react the fastest; they react via S_N1.

(Within a primary/secondary/tertiary series, iodoalkanes react faster than bromoalkanes, that react faster than chloroalkanes because the bonds are longer and weaker. Bond length: C−I > C−Br > C−Cl bond.)

Therefore (CH₃)₃Cl is the correct answer.

Ketones (CH₃CH₂COCH₃ is butanone) cannot be oxidised using K₂Cr₂O₇ / H⁺.

However, they can be reduced to secondary alcohols (in this case butan-2-ol) using NaBH_4(aq) followed by reaction with H⁺.

Therefore 2 only is the correct answer.

This is a Markovnikov electrophilic addition reaction mechanism.

The hydrogen bromide (H−Br) will add across the double bond. CH₃CHBrCH₂Br has two bromine atoms added, and CH₃CH=CHBr still has a double bond, so these are both incorrect answers.

The mechanism will go preferentially via the secondary carbocation (positive inductive effect makes the secondary more stable) and produce 2-bromopropane (CH₃CHBrCH₃) rather than 1-bromopropane (CH₃CH₂CH₂Br) which is formed via the primary carbocation.

Therefore CH₃CHBrCH₃ is the correct answer.

Polar protic solvents can form H-bonds, e.g. water and ethanol, and they favour S_N1, as they are effective at stabilising the carbocation and negative nucleophile. Polar aprotic solvents cannot form H-bonds, e.g. propanone and ethoxyethane, and they favour S_N2.

All the solvent molecules shown here can form H-bonds apart from CH₃COCH₃, propanone.

CH₃COCH₃ is therefore the correct answer.

Benzene is a planar molecule, with the six carbon atoms forming the shape of a regular hexagon. The carbon-carbon bonds are all the same length, but that is an intermediate length between a double and single bond ('bond order' 1.5). Although we sometimes draw it with three double (C=C) and three single (C−C) bonds, it does not possess double bonds because benzene is a delocalised structure. The delocalised structure gives benzene extra stability.

The extra stability means that benzene will not react with Br₂ in an electrophilic addition reaction, since that would disrupt the delocalisation. Benzene will however react with NO₂⁺ in an electrophilic substitution reaction, since the delocalisation is maintained in the product.

Therefore statement 1 and 2 only are correct.

Cis-trans isomerism is shown by compounds when the two substituents attached to both of the carbon atoms of the C=C double bond are different, or when two substituents are on different carbon atoms in a cyclic carbon compound.

1,2-dimethylcyclobutane is shown below:

1,2-dimethylcyclobutane will show cis-trans isomerism, since there is no free rotation around the bonds in the ring, so the two methyl groups can be oriented 'up' from the plane of the ring, or 'down' from the plane of the ring. This diagram shows trans (one 'up' and one 'down').

The other compounds all have two substituents that are the same attached to one of the carbon atoms of the double bond, or have no substituents attached to the ring in the case of cyclobutane.

Therefore 1,2-dimethylcyclobutane is the correct answer.

Primary haloalkanes (1-bromopropane) react via an S_N2 mechanism. Tertiary haloalkanes (2-bromo-2-methylpropane) react via an S_N1 mechanism. So neither statement 1 nor 2 are correct.

Within a primary/secondary/tertiary series, iodoalkanes react faster than bromoalkanes, that react faster than chloroalkanes because the bonds are longer and weaker. Bond length: C−I > C−Br > C−Cl bond. So statement 3 is correct.

Therefore 3 only is the correct answer.

A chiral carbon atom, is a carbon atom with four different substituents attached.

CH₃CBr(CH₃)CH₂CHBrCH₃ is 2,4-dibromo-2-methyl-pentane and is shown below:

The fourth carbon atom in the chain has four different substituents attached and so is chiral. The other carbons are not chiral (any carbon with 2 or more hydrogen atoms attached cannot be chiral and the second carbon has two CH₃ groups attached).

Therefore 1 is the correct answer.

In E/Z isomerism each substituent is given a priority according to the atomic number (Z) of the atom attached to the carbon. Moving to the next atom in the event of a ‘draw’. The highest substituents on each carbon atom are then compared to see if they are E (entgegen) or Z (zusammen) relative to one-another.

In this molecule, across the double bond, the bromine atom (Z=35) has higher priority than chlorine (Z=17) on carbon 1 and the carbon atom (Z=6) has higher priority than hydrogen (Z=1) on carbon 2.

In this molecule, the highest priorities are therefore on the opposite side of the carbon chain and therefore the E, 'entgegen', isomer.

The molecule has four carbons in the longest chain, so the full name is: (E)-1-bromo-1-chlorobut-1-ene which is therefore the correct answer.

Organic reaction types, reagents, catalysts and conditions need to be learned.

Nitration of benzene (C₆H₆) is carried out using concentrated nitric acid and concentrated sulfuric acid in a 'nitrating mixture'. (The sulfuric acid acts as a catalyst, as it is re-produced at the end of the reaction.)

Conversion of nitrobenzene (C₆H₅NO₂) to phenylamine (C₆H₅NH₂) is carried out in a two step process: using tin (Sn) and concentrated hydrochloric acid first, and then in a second step, using sodium hydroxide.

Therefore Step 1 conc. HNO₃ and conc. H₂SO₄; Step 2 Sn and conc. HCl; Step 3 dilute NaOH (aq) is the correct answer.

Organic reaction types, reagents, catalysts and conditions need to be learned.

The substitution of hydrogen atoms on the benzene ring can only be carried out by electrophilic substitution. The nitration of benzene is the example we usually see, but other atoms/groups can be substituted in by the same mechanism.

The substitution of hydrogen atoms on an alkyl group (or alkane) by halogens is carried out by free radical substitution. (Note that the substitution in reaction 2 is on the alkyl chain and not on the benzene ring.)

Therefore Reaction 1: Electrophilic Reaction 2: Free radical is the correct answer.

Organic reaction types, reagents, catalysts and conditions need to be learned.

Reaction 1 is a Markovnikov electrophilic addition reaction. The hydrogen chloride (H−Cl) will add across the double bond. The mechanism will go preferentially via the secondary carbocation (positive inductive effect makes the secondary more stable) and produce 2-chloropropane (CH₃CHClCH₃).

Reaction 2 is a nucleophilic substitution reaction, the chlorine atom will be substituted by the hydroxyl group, forming propan-2-ol (CH₃CHOHCH₃).

Reaction 3 is an oxidation reaction. The secondary alcohol, propan-2-ol, will form a ketone when oxidsed with acidified dichromate ions, that is propanone (CH₃COCH₃).

Therefore CH₃COCH₃ is the correct answer.