MDCAT Chemistry chapter 14: Chemistry · Hydrocarbons MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
55 exam-style questions on Chemistry · Hydrocarbons, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Ten Chemistry · Hydrocarbons questions with answers
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. Removing one hydrogen atom from the CH3 group of methylbenzene (toluene), C6H5CH3, gives which group?
A
benzyl, C6H5CH2–
B
phenyl, C6H5–
C
tolyl, CH3C6H4–
D
benzoyl, C6H5CO–
Show answer
Answer: A. Loss of H from the side chain leaves C6H5CH2–, the benzyl group. The phenyl group, C6H5–, comes from removing H from benzene itself, and removing a ring hydrogen from toluene gives a tolyl group.
2. One mole of bromine is added in the cold to one mole of pent-1-en-4-yne, CH2=CH–CH2–C≡CH. Where does the bromine mainly add, and why?
A
to the C≡C bond, as it has more π electrons than the C=C bond
B
to the C=C bond, as alkenes react faster with electrophiles
C
to the C≡C bond, as its terminal hydrogen is acidic
D
to both bonds equally, as both are unsaturated
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Answer: B. Electrophilic addition is faster at a C=C bond than at a C≡C bond, whose π electrons are held more tightly by sp carbons. So the main product is 4,5-dibromopent-1-yne, with the triple bond left intact.
3. Iodine does not iodinate methane to any useful extent unless an oxidising agent such as HIO3 is present. The oxidising agent is needed because
A
it splits I2 into iodine free radicals even in the dark
B
the HI formed reduces CH3I back to methane, so it must be removed
C
methane must first be oxidised to methanol, which then reacts with iodine
D
it converts methane into methyl free radicals that iodine can attack
Show answer
Answer: B. CH4 + I2 ⇌ CH3I + HI is reversible, and HI is a strong reducing agent that turns CH3I back into CH4. An oxidising agent destroys the HI, so the forward reaction can proceed.
4. The reactivity of the halogens towards alkanes in substitution reactions decreases in the order
A
I2 > Br2 > Cl2 > F2
B
Cl2 > F2 > Br2 > I2
C
F2 > Cl2 > Br2 > I2
D
Cl2 > Br2 > I2 > F2
Show answer
Answer: C. Fluorine reacts explosively with alkanes even in the dark, chlorine and bromine need light or heat, and iodine hardly reacts at all. The order follows the falling reactivity of the halogens down group 17.
5. The IUPAC name of CH≡C–CH2–CH=CH–CH3 is
A
hex-2-en-5-yne
B
hex-1-yn-4-ene
C
hex-4-en-1-yne
D
hex-1-en-4-yne
Show answer
Answer: C. Number the six-carbon chain to give the multiple bonds the lowest set of locants: 1 and 4 from the C≡C end, against 2 and 5 from the other end. The ending is written -en- before -yne, giving hex-4-en-1-yne.
6. Alkanes were once called paraffins (little affinity) because they react with few reagents. The main reason for their low reactivity is that
A
their C–C and C–H σ bonds are strong and almost non-polar
B
their molecules contain weak π bonds that are easily broken
C
their C–H bonds are highly polar and attract water strongly
D
their carbon atoms carry lone pairs that repel most reagents
Show answer
Answer: A. Carbon and hydrogen have similar electronegativities, so the bonds are nearly non-polar and offer no charged site to ionic reagents, and the C–C and C–H σ bonds are strong. Alkanes are saturated and have no π bonds.
7. All six carbon–carbon bonds in benzene have the same length because
A
the ring has three fixed double bonds and three single bonds
B
the π electrons are delocalised over the whole ring
C
each carbon atom is sp3 hybridised
D
the ring is puckered like a chair
Show answer
Answer: B. Each sp2 carbon gives one p electron to a π cloud spread evenly above and below the ring, so every C–C bond is in between a single and a double bond. Fixed alternate double bonds would give two different bond lengths.
8. An alkyne is prepared by heating a vicinal dihalide, such as BrCH2CH2Br, with an excess of a strong base (alcoholic KOH or NaNH2). In this reaction each dihalide molecule loses
A
two molecules of HBr, one after the other
B
one molecule of Br2
C
one molecule of HBr only
D
two Br− ions, which are replaced by OH groups
Show answer
Answer: A. Each elimination removes H from one carbon and Br from the next: BrCH2CH2Br → CH2=CHBr → HC≡CH, a double dehydrohalogenation. Losing Br2 (with zinc dust) gives only an alkene, and replacement by OH groups is substitution.
9. Methane and chlorine react in diffused sunlight to give a mixture of CH3Cl, CH2Cl2, CHCl3 and CCl4. A mixture forms because
A
methane contains four different kinds of hydrogen atom
B
the CH3Cl formed is itself a free radical
C
Cl2 splits into Cl+ and Cl− ions in light
D
Cl• also removes H from chloromethanes already formed
Show answer
Answer: D. Once CH3Cl builds up, chlorine radicals take hydrogen from it as well (CH3Cl + Cl• → CH2Cl• + HCl), and the chain continues to CH2Cl2, CHCl3 and CCl4. All four hydrogen atoms of methane are equivalent, and the products are stable molecules, not radicals.
10. Benzene is first nitrated, and the product is then chlorinated with Cl2 in the presence of FeCl3. The main final product is
A
o-chloronitrobenzene
B
m-chloronitrobenzene
C
p-chloronitrobenzene
D
m-dichlorobenzene
Show answer
Answer: B. Nitration gives nitrobenzene, and its –NO2 group sends the chlorine to the meta position. Chlorinating first and nitrating second would give mainly o- and p-chloronitrobenzene, because –Cl directs ortho/para.
55, each with the answer and a short explanation, across 2 topics: Alkanes and alkenes, Alkynes and benzene. Every chapter test draws ten of them, so you can take it many times.
Is the MDCAT Chemistry chapter 14 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
