7

The major product is the Zaitsev one i.e 1-methylcyclohexene. The OH needs to be protonated first in order for it to leave and moreover it is a very fast leaving group (also we have the case that $\ce{HSO_{4}^{-}}$ or $\ce{H_{2}O}$ are very weak bases and considering that only elimination reactions occur tells us that it must follow an E1 elimination). When ...


2

Many people confuse between the three forms of the OH group. When the OH has a negative charge, it is called hydroxide, and it forms ionic compounds with cations. An example is sodium hydroxide, which contains discrete Na+ ions and OH- ions. When dissolved in water, it separates into Na+ ions and OH- ions (solvation not mentioned for simplicity) When the OH ...


2

It does seem that placing the hydroxyl group at the terminal carbon maximises the hydrogen bonding interactions between alcohol molecules and also the van der Waal forces between them. This is best illustrated with a crude diagram I have made: Finer differences due to the position of the hydroxyl group along the chain, such as the differences between the 2- ...


2

The proposed solution to this roadmap problem has several errors. I am assuming that $\ce{K2Cr2O7}$, methanol and $\ce{LiAlH4}$ are employed in excess. In the oxidation of 1 (A), chromic acid, $\ce{H2Cr2O7}$, is formed from $\ce{K2Cr2O7}$. To obtain the purported hydroxy acid 6 suggested in the solution, exclusive oxidation of the primary alcohol would be ...


1

Welcome on Chemistry SE! Here, you are doing an esterification, done with your carboxylic acid and methanol with $\ce{H+}$. The problem is, the yield for such reaction is quite low (an average 67% for a primary alcohol, because of the equilibrium between esterification and hydrolysis). A solution to increase it is to use an excess of reagant (preferably the ...


1

As other answers point out, the hydroxyl group in an alcohol is covalently bonded to carbon and thus is not the same as a hydroxide ion. But don't assume that the hydroxyl group lacks any basic character. It can still act as a Lewis base towards strong acids like $\ce{HCl}$, facilitating nucleophilic substitution reactions between alcohols and these strong ...


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