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Write the Iupac Name of the Following : - Chemistry

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प्रश्न

Write the IUPAC name of the following :

उत्तर

The parent chain contains 5 carbon atoms. Numbering will start from the right-hand side. Hence, the name of the compound is 3,3-dimethylpentan-2-ol

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2017-2018 (March) Delhi Set 1

वीडियो ट्यूटोरियलVIEW ALL [3]

संबंधित प्रश्न

Write structural formulae for 1-Ethylcyclohexanol.


In a carbinol system of nomenclature tert.butyl alcohol is named as _______________


Isopropyl alcohol on oxidation forms:


Give IUPAC names of the following compound:


The compound HOCH2 – CH2OH is __________.


Which of the following is most acidic?


The correct acidic strength order of the following is:


     (I)


    (II)


     (III)


Match the starting materials given in Column I with the products formed by these (Column II) in the reaction with HI.

  Column I   Column II
(i) CH3—O—CH3 (a)
(ii) \[\begin{array}{cc}
\ce{CH3}\phantom{..................}\\
\backslash\phantom{.............}\\
\ce{CH-O-CH3}\\
/\phantom{..............}\\
\ce{CH3}\phantom{..................}
\end{array}\]
(b) \[\begin{array}{cc}
\ce{CH3}\phantom{....}\\
|\phantom{.......}\\
\ce{CH3-C-I + CH3OH}\\
|\phantom{.......}\\
\ce{CH3}\phantom{....}
\end{array}\]
(iii) \[\begin{array}{cc}
\ce{CH3}\phantom{.}\\
|\phantom{....}\\
\ce{H3C-C-O-CH3}\\
|\phantom{....}\\
\ce{CH3}\phantom{..}
\end{array}\]
(c)
(iv) (d) CH3—OH + CH3—I
    (e) \[\begin{array}{cc}
\ce{CH3}\phantom{.....................}\\
\backslash\phantom{.................}\\
\ce{CH-OH + CH3I}\\
/\phantom{.................}\\
\ce{CH3}\phantom{.....................}
\end{array}\]
    (f) \[\begin{array}{cc}
\ce{CH3}\phantom{.....................}\\
\backslash\phantom{.................}\\
\ce{CH-I + CH3OH}\\
/\phantom{.................}\\
\ce{CH3}\phantom{.....................}
\end{array}\]
    (g) \[\begin{array}{cc}
\ce{CH3}\phantom{....}\\
|\phantom{.......}\\
\ce{CH3-C-OH + CH3I}\\
|\phantom{.......}\\
\ce{CH3}\phantom{....}
\end{array}\]

Assertion: Phenol forms 2, 4, 6 – tribromophenol on treatment with \[\ce{Br2}\] in carbon disulphide at 273 K.

Reason: Bromine polarises in carbon disulphide.


Give the structures of Thiosulphuric acid and Peroxy monosulphuric acid.


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