Answer: Option (d) is the correct answer.
Explanation:
Steps involved for the given reaction will be as follows.
Step 1: 
(fast)
Rate expression for step 1 is as follows.
Rate = k ![[NO]^{2}](https://tex.z-dn.net/?f=%5BNO%5D%5E%7B2%7D)
Step 2: 
This step 2 is a slow step. Hence, it is a rate determining step.
Step 3. 
(fast)
Here, 
is intermediate in nature.
All the steps are bimolecular and it is a second order reaction. Also, there is no catalyst present in this reaction.
Thus, we can conclude that the statement step 1 is the rate determining step, concerning this mechanism is not directly supported by the information provided.
Answer:
1) acetylide
2) enol
3) aldehydes
4) tautomers
5) alkynes
6) Hydroboration
7) Keto
8) methyl ketones
Explanation:
Acetylide anions (R-C≡C^-) is a strong nucleophile. Being a strong nucleophile, we can use it to open up an epoxide ring by SN2 mechanism. The attack of the acetylide ion occurs from the backside of the epoxide ring. It must attack at the less substituted side of the epoxide.
Oxomercuration of alkynes and hydroboration of alkynes are similar reactions in that they both yield carbonyl compounds that often exhibit keto-enol tautomerism.
The equilibrium position may lie towards the Keto form of the compound. Usually, if terminal alkynes are used, the product of the reaction is a methyl ketone.
The correct response would be 3. The alkaline earth metals would tend to lose its valence electrons, in this case 2 of them at different energy levels, to form the same respective ion, which is +2.
D. Pentane. Pentane has 5 Carbon and 12 Hydrogen, like the molecule shown.
