Q.Name the product obtained in the following reaction? CH3-CH2-CH2-CH=CH2 (Pent-1-ene) --(BH3)2 / H2O2/OH^- --> ?

Acid-catalysed hydration of an alkene (H+/H2O) is a Markovnikov addition: the -OH group ends up on the carbon that can best stabilise the intermediate carbocation, i.e. the more substituted carbon.

Mechanism: H+ protonates the double bond of CH3-CH=CH2. Protonation occurs so as to generate the more stable carbocation - here, protonating the terminal CH2 gives a secondary carbocation on the middle carbon, CH3-CH+-CH3, which is more stable than the alternative primary carbocation.

Water then attacks this secondary carbocation, and loss of a proton gives the final alcohol:

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General formula of an alkene = CnH2n.

Alkenes contain one C=C double bond and have the general formula CnH2n (e.g. ethene C2H4, propene C3H6). Al …

The electrophilic attack of H3O+ on an alkene protonates the double bond to form a carbocation (the alcohol is only formed later, after water addition and deprotonation). Answer: carbocation.

Br$_2$ adds across the C=C double bond of ethene (electrophilic addition), giving the vicinal dibromide.

Alkenes are electron-rich due to the $\pi$ bond, so they readily undergo electrophilic addition with bromine. In CCl$_4$ (an inert non-aqueous solvent, used so no other nucleophile interferes), $Br_2$ adds directly across the double bond:

$CH_2=CH_2 + Br_2 \xrightarrow{CCl_4} CH_2Br-CH_2Br$

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Markovnikov addition of water (via a more stable secondary carbocation intermediate) places -OH on the middle carbon of propene, giving 2-propanol.

CH3-CH=CH2 + H2O --(dil. H2SO4)--> CH3-CH(OH)-CH3 …

The pi bond in the C=C double bond of alkenes is weak and electron-rich, so it is easily attacked by electrophiles — this makes alkenes far more reactive than the saturated alkanes.

In an alkene, the doubly-bonded carbons are sp2-hybridised. Each carbon forms three sigma bonds in a plane (120° apart) using sp2 orbitals, and the double bond consists of one sigma bond (head-on sp2-sp2 overlap) plus one pi bond, formed by sideways overlap of the unhybridised p-orbitals on the two carbons.

The pi bond has two key features that make alkenes reactive:

1. It is weaker than a sigma bond (sideways p-orbital overlap is less effective than head-on overlap), so it breaks more easily than a C-C sigma bond.
2. Its electron cloud is spread above and below the molecular plane, away from the nuclei, making these electrons loosely held and easily accessible/polarisable. …

Direct acid-catalysed hydration of propene follows Markovnikov's rule and gives propan-2-ol; to reach the anti-Markovnikov propan-1-ol instead, propene is converted via hydroboration–oxidation.

Why simple hydration doesn't work

Direct acid-catalysed addition of water to propene ($CH_3-CH=CH_2$) follows Markovnikov's rule — $H^+$ adds to the terminal (less substituted) carbon and $OH$ ends up on the more substituted (secondary) carbon — giving propan-2-ol, not the target propan-1-ol.

Hydroboration–oxidation route to propan-1-ol

Step 1 — Hydroboration: diborane ($B_2H_6$, or $BH_3$·THF) adds across the double bond with boron attaching to the less substituted (terminal) carbon (anti-Markovnikov, because it is a concerted, steric/electronic-controlled syn addition where boron preferentially bonds to the less hindered carbon):

$3\,CH_3-CH=CH_2 + B_2H_6 \longrightarrow (CH_3CH_2CH_2)_3B$

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But-1-ene → butane by catalytic hydrogenation, H₂/Pd — option (a).

From NCERT Class 11 Chemistry (Hydrocarbons): an alkene is reduced to an alkane by addition of hydrogen over a metal catalyst (Ni, Pd or Pt):

CH₂=CH–CH₂–CH₃ + H₂ →(Pd) CH₃–CH₂–CH₂–CH₃. …

Acid-catalysed hydration of propene proceeds through the more stable secondary carbocation, so OH ends up on the middle carbon, giving propan-2-ol (Markovnikov addition).

Mechanism

1. Protonation: $H^+$ (from $H_3O^+$) adds to one of the alkene carbons. Protonating the terminal $=CH_2$ carbon generates a secondary carbocation at C-2, $CH_3-\overset{+}{C}H-CH_3$, which is more stable than the alternative primary carbocation that would form if $H^+$ added to the other carbon (Markovnikov's rule: the proton adds to the carbon that already bears more hydrogens, so as to generate the more stable, more substituted carbocation).
2. Nucleophilic attack: water attacks the electrophilic secondary carbocation: $CH_3-\overset{+}{C}H-CH_3 + H_2O \rightarrow CH_3-CH(OH_2^+)-CH_3$. …

Acetylene reacts faster with nucleophiles than ethylene because its sp carbons are more electronegative than ethylene's sp2 carbons.

In ethylene (CH2=CH2), each carbon of the double bond is sp2-hybridised, with 33% s-character. In acetylene (CH≡CH), each carbon of the triple bond is sp-hybridised, with 50% s-character.

Greater s-character means the hybrid orbital electrons (and hence the bonding electrons) are held closer to and more tightly by the nucleus, making sp carbon atoms more electronegative than sp2 carbon atoms. As a result, the carbon atoms of a triple bond are relatively more electron-deficient (carry a greater partial positive character) than those of a double bond, so they attract an electron-rich nucleophile more strongly.

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Direct acid-catalysed hydration of propene would give the Markovnikov (2°) alcohol; to get the anti-Markovnikov, terminal (1°) alcohol, hydroboration–oxidation is used instead.

Simple acid-catalysed addition of water to propene ($CH_3\text{–}CH=CH_2$) follows Markovnikov's rule and would place $-OH$ on the more substituted carbon, giving propan-2-ol — not what is wanted here.

To obtain the terminal alcohol, propan-1-ol, the hydroboration–oxidation sequence is used, which adds $H$ and $OH$ with anti-Markovnikov regiochemistry (boron, and hence eventually $OH$, ends up on the less substituted, terminal carbon):

Step 1 (hydroboration): propene reacts with diborane; boron adds to the less hindered (terminal) carbon: …

Rate ∝ stability of the intermediate cyclopropylcarbinyl cation → I < II < III.

Acid-catalysed (Markovnikov) hydration proceeds via protonation to the most stable carbocation, which here is always on the carbon next to the cyclopropyl ring (cyclopropyl strongly stabilises an adjacent + charge):

• I (cyclopropyl-CH=CH₂): gives a 2° cyclopropylcarbinyl cation. …