Chemistry · Hydrocarbons · NEET
| C-C bond type | Alkane: single | Alkene: double | Alkyne: triple | Arene: delocalised ring |
| General formula | Alkane CnH2n+2 | Alkene CnH2n | Alkyne CnH2n-2 | Arene C6H6 |
| Hybridisation | Alkane sp3 | Alkene sp2 | Alkyne sp | Arene sp2 |
| Bromine water test | Alkane: no change | Alkene: decolourizes | Alkyne: decolourizes | Arene: no change (no catalyst) |
| Typical reaction | Alkane: substitution | Alkene: addition | Alkyne: addition | Arene: substitution |
The difference is the carbon-carbon bond. Alkanes have only single bonds (C-C). Alkenes have at least one double bond (C=C). Alkynes have at least one triple bond (C≡C). Aromatic hydrocarbons (arenes) have a ring of carbons with special shared electrons, like benzene. So the bond type is the first thing to check in NEET.
For an open-chain compound with n carbons: alkane is CnH2n+2, alkene is CnH2n (one double bond), alkyne is CnH2n-2 (one triple bond). Each extra bond removes 2 hydrogens. Benzene is C6H6 (arene, formula CnH2n-6 for the simplest ring). NEET often gives a formula and asks the family, so learn these.
Saturated means the carbons hold the maximum possible hydrogens. Alkanes have only single bonds, so every carbon is full with hydrogen. Alkenes, alkynes and arenes have double or triple or ring bonds, so they carry fewer hydrogens. That empty space is why they are called unsaturated and why they add extra atoms in reactions.
For addition reactions, the order is usually alkyne and alkene much more reactive than alkane. Alkanes are quite unreactive because single bonds are strong and hard to break. Alkenes and alkynes have weak pi bonds that open easily, so they add bromine, hydrogen and acids fast. This is why alkanes mostly do substitution, not addition.
Alkanes and normal aromatic rings like cyclohexane or benzene do NOT decolourize bromine water quickly, because they have no free double or triple bond to react. Alkenes and alkynes DO decolourize it (the brown colour fades) because the pi bond adds bromine. NEET uses this test to tell saturated from unsaturated compounds.
An alkene has a real double bond that reacts fast by addition. An arene like benzene has its pi electrons shared all around the ring (delocalised). This makes benzene extra stable, so it prefers substitution and does not add bromine easily like an alkene does. So arenes look unsaturated but behave differently.
In alkanes each carbon is sp3 (four single bonds). In alkenes the double-bonded carbons are sp2. In alkynes the triple-bonded carbons are sp. In benzene each ring carbon is sp2. NEET links this to bond angle: sp3 is 109.5°, sp2 is 120°, sp is 180° (linear).
The compound that will react most readily with gaseous bromine has the formula:
Which one of the following compounds does NOT decolourize bromine water?
Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Count the bonds between carbons. One line means alkane (single bond). Two lines means alkene (double bond). Three lines means alkyne (triple bond). A six-carbon ring with shared electrons means an aromatic hydrocarbon like benzene.
Aromatic hydrocarbons are technically unsaturated because the ring has double bonds. But the electrons are spread around the whole ring, making benzene very stable. So it does not react by addition like a normal alkene; it prefers substitution.
Yes, but only under sunlight or UV light, and it is a substitution reaction, not addition. A hydrogen is replaced by bromine. This is slow compared to alkenes and alkynes, which add bromine instantly and decolourize bromine water.
A single bond (sigma) is strong and hard to break, so alkanes are unreactive. Double and triple bonds contain weak pi bonds that break easily, so alkenes and alkynes are more reactive in addition reactions even though a triple bond is stronger overall.
Yes. NEET regularly asks which hydrocarbon reacts with bromine, which decolourizes bromine water, hybridisation of carbons, and general formulas. Knowing the four families clearly helps you solve many Hydrocarbons questions quickly.