Chemistry · Aldehydes, Ketones And Carboxylic Acid · NEET
The C=C double bond breaks in the middle. Each carbon that was part of the double bond now becomes a carbonyl carbon (C=O). So one alkene gives two carbonyl compounds (or one if the molecule is symmetric). Rule: look at each doubly-bonded carbon. If it has an H attached, that end becomes an ALDEHYDE. If it has two carbons attached (no H), that end becomes a KETONE. Example: CH3-CH=CH-CH3 gives two molecules of CH3CHO (ethanal).
No. Ozone first makes an unstable ozonide. Zinc dust and water (reductive workup) break that ozonide gently to give aldehydes and ketones. If you skip Zn and use water alone (oxidative workup), the aldehydes get over-oxidised to carboxylic acids. For NEET, the standard reagent set is O3 followed by Zn/H2O and the products are aldehydes/ketones, NOT acids. This is a very common trap.
Only ethyne (HC≡CH, the smallest alkyne) gives an aldehyde, which is acetaldehyde (CH3CHO). Every OTHER alkyne gives a ketone. The reagents are water with H2SO4 and HgSO4 (mercuric sulphate as catalyst). Reason: water adds by Markovnikov rule to form an unstable enol, which rearranges (tautomerises) to the carbonyl. For ethyne both carbons are equal, so you get an aldehyde; for higher alkynes the OH lands on the more substituted carbon, giving a ketone.
Ozonolysis starts from an alkene (C=C) and BREAKS the molecule into two carbonyl pieces. Alkyne hydration starts from an alkyne (C≡C) and does NOT break the molecule; it just ADDS water across the triple bond to give ONE carbonyl compound with the same number of carbons. So ozonolysis = cut and count two products; hydration = add water and keep one product.
Step 1: Draw the alkene and find the C=C. Step 2: Cut it in half; put =O on each carbon that was in the double bond. Step 3: For each new carbonyl carbon, if it still has an H, it is an aldehyde; if it has two carbon groups, it is a ketone. Symmetric alkenes give one product (double amount); unsymmetrical alkenes give two different products.
Match List-I with List-II and choose the correct option. List-I (Reaction): (a) C6H6 -> C6H5-CO-C6H5 (benzophenone); (b) Alkene -> carbonyl (oxidative cleavage); (c) C6H5-OH -> carbonyl (oxidation); (d) C6H5-CH2-CH3 -> C6H5-COOH. List-II (Reagents/Condition): (i) C6H5COCl, anhyd. AlCl3; (ii) CrO3; (iii) KMnO4/KOH, heat; (iv) (1) O3, (2) Zn-H2O.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. Ethene (CH2=CH2) is symmetric, so cutting the C=C gives two molecules of methanal (formaldehyde, HCHO).
Water is added in the presence of dilute H2SO4 with HgSO4 (mercuric sulphate) acting as the catalyst. Ethyne then gives acetaldehyde; higher alkynes give ketones.
The first product is an enol (C=C with an OH). Enols are unstable and rearrange by keto-enol tautomerism to the more stable keto form, which is the aldehyde or ketone.
It is an oxidative cleavage. Ozone adds across the double bond first, but the final result breaks the molecule into two carbonyl fragments, so it counts as cleavage for NEET.
No. Only terminal ethyne gives an aldehyde. Symmetric internal alkynes like but-2-yne give a single ketone; the products are always ketones for any alkyne except ethyne.
