चिकित्सा शिक्षा आयोग (CEE) बीडीएस प्रवेश परीक्षा तयारी
चिकित्सा शिक्षा आयोग (CEE) – बीडीएस प्रवेश परीक्षा
CEE – BDS Entrance · Medical Education Commission – CEE
२,०३९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
MBBS/BDS/Nursing प्रवेश: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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Which of the following compounds gives a positive iodoform test?
सही उत्तर: घ) Ethanol
Ethanol is oxidised to acetaldehyde (CH₃CHO), which has the CH₃–C=O group and gives yellow iodoform (CHI₃). Methanol lacks this CH₃–CH(OH)– unit and does not respond.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Ozonolysis of ethene followed by reductive work-up (Zn/H₂O) gives:
सही उत्तर: ग) Formaldehyde
Ozonolysis cleaves the C=C bond and converts each CH₂ unit into a C=O group, so ethene gives two molecules of HCHO. Ozonolysis of but-2-ene would give acetaldehyde.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Heating sodium acetate with soda lime (NaOH + CaO) produces:
सही उत्तर: ख) Methane
This decarboxylation removes CO₂ (as Na₂CO₃), giving an alkane with one carbon fewer: CH₃COONa → CH₄. Ethane is the product of Kolbe's electrolysis of sodium acetate.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Kolbe's electrolysis of an aqueous solution of sodium acetate gives:
सही उत्तर: ग) Ethane
At the anode, acetate ions lose CO₂ to form methyl radicals, which pair up to give ethane (CH₃–CH₃). Methane is obtained instead by heating sodium acetate with soda lime.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In the nitration of benzene with a mixture of conc. HNO₃ and conc. H₂SO₄, the electrophile is:
सही उत्तर: क) NO₂⁺
H₂SO₄ protonates HNO₃, which loses water to form the nitronium ion NO₂⁺, the attacking electrophile. NO₂⁻ (nitrite) is negatively charged and cannot act as an electrophile.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Alkyl halides react with sodium metal in dry ether to form higher alkanes. This reaction is called:
सही उत्तर: क) Wurtz reaction
In the Wurtz reaction, 2R–X + 2Na → R–R + 2NaX, e.g. CH₃Br gives ethane. The Fittig reaction is the analogous coupling for aryl halides.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The peroxide effect (anti-Markovnikov addition) is observed with:
सही उत्तर: क) HBr only
Only with HBr are both steps of the free-radical chain energetically favourable. With HCl the H–Cl bond is too strong, and with HI the I· radical is too unreactive to add to the alkene.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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The major product of the addition of HBr to propene (in the absence of peroxide) is:
सही उत्तर: घ) 2-Bromopropane
By Markovnikov's rule, H adds to the carbon with more hydrogens, via the more stable secondary carbocation, giving 2-bromopropane. In the presence of peroxide, 1-bromopropane forms (anti-Markovnikov).
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which of the following is the strongest acid?
सही उत्तर: घ) Trichloroacetic acid
Three electron-withdrawing Cl atoms stabilise the carboxylate ion most through the −I effect, making trichloroacetic acid the strongest. Acetic acid, with no Cl, is the weakest of these.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Which of the following compounds is optically active?
सही उत्तर: घ) Lactic acid (2-hydroxypropanoic acid)
The central carbon of lactic acid carries four different groups (H, OH, CH₃, COOH), making it chiral. Glycerol has no carbon with four different groups.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
यी अभ्यास प्रश्न हुन्, आधिकारिक प्रश्न होइनन्। पाठ्यक्रम, विज्ञापन र नतिजाका लागि सम्बन्धित निकायको आधिकारिक वेबसाइट हेर्नुहोस्। उत्तरमा गल्ती भेटे प्रश्नमुनिको "टिप्पणी / गल्ती रिपोर्ट" मा लेख्नुहोस्।
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