
How can you prepare alkoxide?
Answer
536.7k+ views
Hint: An alkoxide is basically defined as the conjugate base of an alcohol and hence it contains an organic group which is bonded to a negatively charged oxygen atom. They are represented by the formula $R{{O}^{-}}$where R is basically an organic substituent. There are various methods for the preparation of alkoxides. Let us discuss them.
Complete step-by-step answer: Alkoxides are generally strong bases and if the substituent attached with them is not bulky then they even behave as good nucleophiles. These alkoxides can be prepared by various methods. But there are two major methods through which most of their production occurs.
1. By reducing metals and alcohol reactions
2. By electrochemical process.
Let us see these two methods one by one.
1. By reducing metals and alcohol reactions:
Alkoxides can be prepared through various steps starting from an alcohol.
In general, the reaction occurs as follows:
Highly reducing metals such as sodium reacts directly with alcohols to produce the corresponding metal alkoxide. We should note that in such reactions, alcohol serves as an acid whereas hydrogen is produced as by-product.
A classic example is preparation of sodium methoxide by the addition of Na metal to methanol. The reaction for the same is represented below:
\[\text{2}C{{H}_{3}}OH+2Na\to 2C{{H}_{3}}ONa+{{H}_{2}}\]
We should note that we can use other alkali metals beside sodium and other alcohols can be used instead of methanol.
Another reaction similar as above occurs when we react alcohol with a metal hydride (for example $NaH$). In this reaction we should note that metal hydride removes hydrogen atom from hydroxyl group and in turn forms an alkoxide ion which is negatively charged.
2. By electrochemical processes.
We can prepare many alkoxides by anodic dissolution of the corresponding metals alcohols which are water-free and the reaction occurs in the presence of electroconductive additive.
The metals which are used for this reaction may be Co, Ge, Mo, Re, Ta, W, Zr, Y, Nb etc.
In the reaction the conductive additive thus used may be either lithium chloride or quaternary ammonium halide.
Some examples of few metal alkoxides which are obtained by this technique are as follows
\[Ti{{(OCH{{(C{{H}_{3}})}_{2}})}_{4}}, \text{ }T{{a}_{2}}{{(OC{{H}_{3}})}_{10}}, \text{ }{{[MoO{{(OC{{H}_{3}})}_{4}}]}_{2}}, \text{ }R{{e}_{2}}{{O}_{3}}{{(OC{{H}_{3}})}_{6}}\text{ }\]
Note:The alkoxide ion can easily react with a primary alkyl halide through the substitution bimolecular reaction mechanism to form an ether. This famous reaction is termed as the Williamsons- ether synthesis and is also the major source for the production of ethers.
Complete step-by-step answer: Alkoxides are generally strong bases and if the substituent attached with them is not bulky then they even behave as good nucleophiles. These alkoxides can be prepared by various methods. But there are two major methods through which most of their production occurs.
1. By reducing metals and alcohol reactions
2. By electrochemical process.
Let us see these two methods one by one.
1. By reducing metals and alcohol reactions:
Alkoxides can be prepared through various steps starting from an alcohol.
In general, the reaction occurs as follows:
Highly reducing metals such as sodium reacts directly with alcohols to produce the corresponding metal alkoxide. We should note that in such reactions, alcohol serves as an acid whereas hydrogen is produced as by-product.
A classic example is preparation of sodium methoxide by the addition of Na metal to methanol. The reaction for the same is represented below:
\[\text{2}C{{H}_{3}}OH+2Na\to 2C{{H}_{3}}ONa+{{H}_{2}}\]
We should note that we can use other alkali metals beside sodium and other alcohols can be used instead of methanol.
Another reaction similar as above occurs when we react alcohol with a metal hydride (for example $NaH$). In this reaction we should note that metal hydride removes hydrogen atom from hydroxyl group and in turn forms an alkoxide ion which is negatively charged.
2. By electrochemical processes.
We can prepare many alkoxides by anodic dissolution of the corresponding metals alcohols which are water-free and the reaction occurs in the presence of electroconductive additive.
The metals which are used for this reaction may be Co, Ge, Mo, Re, Ta, W, Zr, Y, Nb etc.
In the reaction the conductive additive thus used may be either lithium chloride or quaternary ammonium halide.
Some examples of few metal alkoxides which are obtained by this technique are as follows
\[Ti{{(OCH{{(C{{H}_{3}})}_{2}})}_{4}}, \text{ }T{{a}_{2}}{{(OC{{H}_{3}})}_{10}}, \text{ }{{[MoO{{(OC{{H}_{3}})}_{4}}]}_{2}}, \text{ }R{{e}_{2}}{{O}_{3}}{{(OC{{H}_{3}})}_{6}}\text{ }\]
Note:The alkoxide ion can easily react with a primary alkyl halide through the substitution bimolecular reaction mechanism to form an ether. This famous reaction is termed as the Williamsons- ether synthesis and is also the major source for the production of ethers.
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